Demystifying Multivariable Calculus!

Partial Derivatives - Multivariable Calculus

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Summary

In this educational video, The Organic Chemistry Tutor dives deep into the world of partial derivatives in multivariable calculus. Starting with the basics, the tutor outlines how to determine partial derivatives with respect to different variables, using example functions to illustrate the process. Viewers are guided through derivative calculations of increasingly complex functions, incorporating exponential, logarithmic, and trigonometric components. The video also covers higher-order partial derivatives and the interplay between them, offering a comprehensive look at how these concepts are applied practically.

Highlights

Learn how to find partial derivatives of multivariable functions step-by-step! 📚
Understand treating other variables as constants when taking partial derivatives. 🔍
Master the power rule for derivatives and see it in action with examples. ⚡
Find out how to differentiate exponential functions and apply the chain rule. 🚀
Explore methods for differentiating logarithmic functions with precision. 📏
Enhance your skills with derivative applications in trigonometric contexts. 🎲
Discover higher order partial derivatives and understand their notations! 🔢
Grasp the mixed derivative theorem for continuous functions! 🤓
Master practical examples of applying product and quotient rules when needed. ✅
Practice evaluating derivatives at specific points for real-world applications. 🕵️‍♂️

Key Takeaways

Partial derivatives are used to find the derivative of functions with multiple variables, focusing on one variable at a time while treating others as constants. 🤔
Applying the power rule and recognizing when to treat other variables as constants is crucial in partial differentiation. ✍️
Higher order partial derivatives are derivatives of derivatives, providing a deeper understanding of the changes in multivariable functions. 🔍
Utilizing product and quotient rules appropriately when variables appear in both the numerators and denominators of functions is essential. 🤓
Mixed partial derivatives, where the order of differentiation does not matter, offer insights into specific critical points of the functions. 🔄

Overview

Partial derivatives are a key concept in multivariable calculus, allowing us to explore the behavior of functions with several variables. In this video by The Organic Chemistry Tutor, we embark on a journey through this fascinating topic. By learning how to take derivatives with respect to one variable while treating others as constants, viewers gain insight into this crucial mathematical tool.

Through a series of in-depth examples, the video demonstrates how to apply the power rule in partial differentiation, and highlights the usefulness of chain, product, and quotient rules. From simple polynomial functions to those involving exponential and trigonometric expressions, the tutor skillfully navigates these waters, ensuring a comprehensive understanding for the audience.

The exploration does not end there; higher order partial derivatives and the equality of mixed derivatives are also discussed. These advanced concepts expand the viewer's skillset, enabling them to tackle higher-level problems and understand continuous functions more profoundly. This practical approach, combined with examples involving the evaluation of functions at given points, cements the real-world utility of these concepts.

Chapters

00:00 - 01:00: Introduction to Partial Derivatives In the 'Introduction to Partial Derivatives' chapter, the focus is on finding partial derivatives with respect to a certain variable. The explanation begins with considering a function that has multiple variables, such as X and Y, and illustrates the concept using an example function (7x^2 - x^3y^4 + 5x^4y).
01:00 - 02:30: Power Rule Review The chapter 'Power Rule Review' discusses how to find the partial derivative of a function with respect to X. It emphasizes treating all other variables (like Y) as constants while considering X as the variable. The chapter proceeds with a review of the power rule before solving a problem related to this concept.
02:30 - 03:30: Partial Derivative with Respect to X The chapter titled 'Partial Derivative with Respect to X' explains how to find the derivative of functions of the form x raised to the power of N. According to the transcript, the derivative of such a function is calculated as N times x raised to the power of N minus one. Several examples are provided: the derivative of x^4 is 4x^3, x^3 is 3x^2, x^5 is 5x^4, x^2 is 2x, and x^1 is 1x. The concept covered is fundamental in calculus and essential for understanding how to compute derivatives.
03:30 - 05:00: Partial Derivative with Respect to Y The chapter discusses the concept of taking partial derivatives with respect to the variable Y. It begins with a review of basic derivatives, such as the derivative of x^2, which is 2x, and the derivative of x^3, which is 3x^2. In this context, the variable Y is treated as a constant. The discussion then proceeds to the derivative of x^4, which is 4x^3, emphasizing that the presence of the constant Y does not change in the differentiation process.
05:00 - 06:30: Practice Example 1 This chapter explains the concept of finding the first partial derivative with respect to a specific variable. It provides an example where X is treated as the variable while other variables are constants. The chapter concludes with the calculated first partial derivative of the given expression, resulting in 14x - 3x^2y^4 + 20x^3y.
06:30 - 10:00: Exponential Function Derivatives This chapter discusses the derivatives with respect to multiple variables, particularly focusing on how to differentiate terms involving both X and Y variables. It begins with the concept of treating certain variables as constants and details the procedure exemplified through expressions such as 7x^2 and y^4. In differentiating with respect to Y, terms including X are considered unchanging; hence their derivatives are plotted along alterations in Y. It covers the step-by-step approach to compute these derivatives, noting for instance, that while differentiating y^4 with respect to Y results in 4y^3, and y^3 becomes 3y^2 when differentiated. This establishes foundational techniques in handling multi-variable calculus, illustrating how each component is impacted as you derive in respect of another variable.
10:00 - 15:00: Partial Derivative of Natural Log Functions The transcript discusses the calculation of partial derivatives of natural log functions. It begins with the derivative of some constant expressions and then provides another example for practice. In this example, the function Z is given by Z = 3x^2 y^4 - 5x^7. The focus is on finding the partial derivative with respect to y.
15:00 - 19:00: Partial Derivative of Logarithmic Function Example The chapter focuses on finding the first partial derivative of a logarithmic function. It begins with establishing that z is a function of x and y, depicted as f(x,y) or simply z. To find the derivative with respect to x, y is treated as a constant, and differentiation is performed only with respect to x. This sets the stage for further exploration into differentiating with respect to y.
19:00 - 20:30: Partial Derivative of Square Root Functions The chapter discusses the process of finding the partial derivatives of square root functions. It elaborates on differentiating terms with variables such as x^2 and x^7, where the derivatives are 2X and 7x^6, respectively. It also touches upon terms without the variable x, like 4y^8, treating them as constants, which results in their derivatives being zero. The final discussed expression for the first partial derivative ends up being 6xy^4 - 35x^6.
20:30 - 23:30: Partial Derivative of Trigonometric Functions In this chapter, the focus is on calculating the partial derivatives of trigonometric functions, specifically with respect to X and Y. The transcript details the process of treating X as a constant while finding the derivative with respect to Y. It explains the derivative of y to the power of 4 as 4y cubed and identifies constants like 5x to the power of 7, which become zero. Further, it calculates the derivative of y to the power of 8 as 8y to the power of 7 and completes the partial differentiation with constants such as 3 times 4 equaling 12 and 4 times 8 equaling 32.
23:30 - 30:00: Evaluating Partial Derivatives at a Point The chapter discusses evaluating partial derivatives at a specific point, particularly focusing on an exponential function where Z equals e to the power of x squared times y. It involves calculating the partial derivatives with respect to X and Y, denoted as dz/dx and dz/dy. The function Z is also referred to as F of x and y in the explanation.
30:00 - 35:00: Partial Derivatives of Functions with Three Variables This chapter discusses partial derivatives of functions with three variables. It covers the notation used to denote partial derivatives, specifically with respect to variables X and Y, using symbols like F subx and Fy. The chapter highlights the interpretation of these symbols and explores the concept of taking derivatives of exponential functions, focusing on the derivative of e^u which includes the term e^u * U'.
35:00 - 47:00: Higher Order Partial Derivatives In the chapter titled 'Higher Order Partial Derivatives,' the discussion began with explaining how to find the derivative of exponential functions involving polynomials. Using e to the power of a polynomial, like e^(x^3), requires multiplying by the derivative of that polynomial (in this case, 3x^2). Similarly, for e^(5x), the derivative involves multiplying by 5, and for e^x, it involves multiplying by 1. Examples were provided to illustrate how to apply the formula for these situations.
47:00 - 55:00: Mixed Partial Derivatives and Theorem The chapter introduces the concept of mixed partial derivatives, applying this to a specific problem example involving exponential and polynomial expressions. It guides the reader through finding partial derivatives dz/dx, maintaining the primary formula for reference.
55:00 - 60:00: Conclusion The chapter discusses finding the partial derivative of a function involving x and y. Specifically, it focuses on treating y as a constant and differentiating x^2, resulting in 2x, while keeping the term y^b unchanged. The process avoids using the product rule as the differentiation is solely with respect to x. The final expression involves 2x multiplied by y^b and multiplied by e raised to the power of x^2y^b.

Partial Derivatives - Multivariable Calculus Transcription

00:00 - 00:30 in this video we're going to talk about how to find the partial derivative with respect to a certain variable so for example let's say if we have a function that consists of multiple variables in this case X and Y and let's say it's equal to 7 x^2 minus X Cub y 4 plus 5x 4 YB so what is the part paral derivative
00:30 - 01:00 of this function with respect to X how can we find the answer the partial derivative with respect to X is symbolized by the variable F subx now in order to find it every other variable in this case y you need to treat it as a constant you only treat X as the variable everything else you view it as a constant now before we do this problem let's review the power room let's say if
01:00 - 01:30 we have a function x to the N if we wish to find the derivative it's going to be NX raised to the N minus one so for example the derivative of x 4th is 4X Cub the derivative of x Cub is 3x^2 the derivative of x 5th is 5 x 4th the derivative of x^2 is 2x to the 1 power the derivative of x or x to the first power is 1x to Z which is the same
01:30 - 02:00 as one so what is the derivative of x^2 the derivative of x^2 is 2x now what about the derivative of x Cub we know it's 3x^2 now y we're going to treat it as a constant so we're just going to rewrite it the derivative of x to the 4th is 4X cub and then we won't change the y 3 so
02:00 - 02:30 that's how you could find the first derivative the first partial derivative with respect to X treat X as a variable and any other variable treat it as a constant so this is equal to 14x - 3x^2 y 4 + 20 x Cub y the 3 so that's the answer that's the first partial derivative with respect to
02:30 - 03:00 X now what about the partial derivative with respect to Y what is the answer in that case so 7 x^2 is treated as a constant the derivative of a constant is zero the derivative of y 4th is 4 y cub and the derivative of y 3 is 3 y^2 so in this case we need to treat X as if
03:00 - 03:30 it's a constant so this is going to be -4 x yb+ 15 x 4 y^2 so that's the partial derivative with respect to Y let's try another example for the sake of practice let's say Z is equal to 3x^2 y 4th - 5x 7th
03:30 - 04:00 plus 4 y 8 find the first partial derivative with respect to X and with respect to Y so if you see z z is basically a function of X and Y so you can write it as F ofx comma y or just Z it doesn't matter now the first derivative with respect to X is going to be so keep in mind let's treat y as a constant so differentiate only the X
04:00 - 04:30 variables the derivative of x^2 is 2X and the derivative of x to the 7th is 7 x 6 now in the last term there is no X variable so 4 y 8 is treated as a constant and the derivative of any constant is zero so this is going to be 6 x y 4 - 35 x 6 so that's the first partial derivative
04:30 - 05:00 with respect to X now let's find it with respect to Y so let's treat X as a constant so the derivative of y 4th is 4 y Cub 5x 7th is a constant so that's going to become zero and a derivative of y 8 is 8 Y 7th 3 * 4 is 12 and 4 * 8 is 32
05:00 - 05:30 so this is the answer that's the partial derivative with respect to Y now let's work on an exponential function let's say Z is equal to Eed to x^2 * YB what is DZ DX and DZ Dy so as you recall Z is the same as F ofx comma y so DZ DX is the partial
05:30 - 06:00 derivative with respect to X that's the same as F subx and DZ Dy is the partial derivative of the function Z with respect to Y which is the same as FY so if you see those symbols uh you know what they mean so now let's talk about the derivative of an exponential function the derivative of e to the U is equal to e u * U Prime so let's say
06:00 - 06:30 if we want to find the derivative of e x Cub it's going to be the same thing e to X Cub times the derivative of x Cub which is 3x^2 so if you want to find the derivative of e 5x it's going to be e5x * 5 and E to X is going to be the same thing times the derivative of x which is one so I want to give you some examples so you know how to use this formula
06:30 - 07:00 now let's apply that to the problem that we have let's just get rid of this portion let's keep the formula there so let's find dzdx so it's going to be e to U which is e to the x^2 y Cub it's going to be the original expression times the derivative
07:00 - 07:30 of x^2 YB now we want to find a partial derivative that means we want to treat y as a constant so we're going to differentiate x^2 which is 2X and rewrite y cub and that will be it we don't have to use a product rule here we're just only differentiating the X variable and so that's it it's going to be 2x YB * e raed to the x^2 YB
07:30 - 08:00 now let's find DZ Dy so if Z is equal to e x^2 y Cub let's differentiate both sides with respect to Y so you can represent it using the symbol D over Dy so on the left we're going to have DZ Dy and on right it's
08:00 - 08:30 going to be the same thing e to the U which is e to x^2 YB time the derivative of x^2 YB so we're going to treat X2 as a constant so we're going to rewrite it and multiply by the derivative of y CU which is 3 y^2 so the answer is 3 x^2 y^2 e to the x^2 y Cub so that's DZ Dy which is the same as F
08:30 - 09:00 suby so here's a problem that you could try for practice let's say Z is equal to X 4th e raised to the Y 5 find the first partial derivative with respect to X and with respect to Y so go ahead take a minute and pause the video so let's find dzdx so notice that X to 4th is the variable and e y fifth is the constant
09:00 - 09:30 because it doesn't have an X variable it only has a y variable so we're going to differentiate X 4th which is 4X cub and then simply rewrite the constant so that's dzdx now dzd y x to 4th is now the constant so we're just going to rewrite it and then differentiate e y the 5th so using this formula it's going to be the same thing e ra the Y 5 times the derivative of y 5
09:30 - 10:00 which is 5 y 4 and so these are the two answers now let's say that F ofx comma Y is equal to the natural log of x^2 + y^2 go ahead and find F ofx the partial derivative with respect to X and FY
10:00 - 10:30 so first you need to know how to differentiate natural log functions the derivative of Ln U is equal to U Prime / U so for example if we want to find the derivative of Ln x^2 + 8 it's going to be the derivative of x^2 + 8 which is 2x / whatever stuff was on the inside of the natural log function which is x^2 + 8 so let's say if we want to find the
10:30 - 11:00 derivative of x Cub I mean l and X Cub + 5x - 9 it's going to be the derivative of the inside function which is 3x^2 + 5 divided by the original of the inside function so for example if you want to find the derivative of lnx it's going to be the derivative of x which is one divid the original stuff on the ins which is X so you get
11:00 - 11:30 1X so now let's solve the problem that we have let's find f of x so it's going to be the derivative of the stuff inside that is the derivative of x^2 + y^2 the derivative of x^2 is 2X and y^2 we're going to treat it as a constant because we're trying to find the partial derivative with respect to X so y^2 is 0 ID the original X2 + y^2 so
11:30 - 12:00 it's simply 2x over x^2 + y^2 now if you want to find F suby it's going to be the derivative of y^2 which is 2 y x^2 is a constant so that becomes zero divided by the original X2 + y^2 so this is the first partial derivative with respect to Y and this is the first partial derivative with respect to X here's one you could try it might be a little challenging but go ah go ahead and try it let's say Z is the natural
12:00 - 12:30 log of x^2 / y go ahead and find dzdx and DZ Dy so the derivative of the natural log of U we know it's U Prime over U so U is basically the function on the inside which is x^2 / Y which I'm going to write it as x^2 * 1
12:30 - 13:00 y so now what's U prime it all depends on if we're looking for dzdx or dzd Y let's find dzdx first so U Prime which I'm going to write it as U Prime with respect to X we're going to treat X as a variable so the derivative of x^2 is 2x time the con 1 y we're going to treat 1
13:00 - 13:30 y as a constant so now dzdx is going to be the same as U Prime u u Prime is 2x * 1 over y / U which is x^2 * 1 y so we can cancel 1 Y and we can cancel an X so we're left over with 2 / X so that's the answer uh for the partial derivative
13:30 - 14:00 with respect to X now let's find it with respect to Y so let's find U Prime with respect to Y so x^2 is going to be treated as a constant this time so we can rewrite it now 1 / Y is yus1 and a derivative of Yus 1 is -1 Yus
14:00 - 14:30 2 so U Prime with respect to Y is x^2 * 1/ y^2 so we're going to write it like that so using this formula U Prime over u d zdy is going to be U Prime which isx2 * 1 over y^2 ID the original U value which is x^2 Time 1 over
14:30 - 15:00 y so we can get rid of the x s term and let's multiply the top and the Bottom by y^2 to simplify the complex fraction so these will cancel so on top we're just going to have a negative 1 on the bottom y^2 / Y is y so this is the first partial derivative with respect to Y it's -1 over y now let's try a problem involving
15:00 - 15:30 square roots so let's say f is equal to theare root of X2 + y^2 find a partial derivative with respect to X and Y so first let's rewrite it so this is equal to x^2 + y^2 raised to the 12 so let's find a partial derivative with respect to X so we need to bring the 1/2 to the front and we're going to keep everything on the side the same so basically we need
15:30 - 16:00 to use the chain rule so to speak now 12 minus 1 that's 12 and now we need to find the derivative of the inside function with respect to X the derivative of x^2 is 2X and for y^2 we're going to treat that as a constant so that's Zero 12 * 2 is 1 so therefore the final answer we can also bring this down because it's a negative2
16:00 - 16:30 so the final answer is going to be X on top / x^2 + y^2 raised to the positive 1 12 when you move it from the top to the bottom the exponent will change sign it's going to change from negative to positive so you can write the final answer as x / theare < TK of x^2 + y^2 so that's the first partial derivative with respect to X now what about with respect to
16:30 - 17:00 Y so we're going to take the exponent and move it to the front and keep the inside the same and then subtract 12 by one which will give us negative a half again and now we need to differentiate the inside function X2 will now be treated as a constant so that's zero and y^2 is 2 y so as you can see everything is going to be the same but we're going to have a y variable on top instead of an X variable so this is the final answer it's y over
17:00 - 17:30 the < TK of x^2 + y^2 now let's try a trig function let's say Z is equal to sin XB y 5 find dzdx and DZ to Y so if you want to find the derivative of a sign function the derivative of sin uu is cosine U * U Prime so if we want to differentiate sin x^2 it's going to be cine
17:30 - 18:00 x^2 times the derivative of the inside part the derivative of x^2 is 2x so let's say if you have s x 5th it's going to be cosine X 5times the derivative of x 5 which is 5 x 4th so now let's work on a problem that we have so let's find
18:00 - 18:30 dzdx the derivative of s is cosine and whatever is inside of s we're going to keep it the same we're not going to change it so it's going to be X Cub y 5th and now we need to differentiate X Cub y 5th with respect to X so we're going to treat X as the variable y as the constant the derivative of x Cub is 3x^2 and then we're going to multiply by the constant so that's the answer which we can write it as 3x^2 y 5th and then cosine X Cub y
18:30 - 19:00 5 now let's find DZ Dy so it's going to be cosine X Cub y the 5th the first part is not going to change but the second part will now X Cub is the constant y 5th is the variable so we can rewrite the constant times the derivative of y 5th which is 5 y
19:00 - 19:30 4 so that's 5x Cub y 4 * cosine X Cub y 5th and so that's going to be the answer now let's say if we're given a function f ofx comma Y which is 2x Cub y^ 2 + 5 YB + 4x2 and now we need to evaluate F ofx
19:30 - 20:00 and F of Y at the point 1 comma 2 how can we do so well first let's find a partial derivative with respect to X is going to be the derivative of x Cub which is 3x^2 we're going to treat y^2 as a constant the derivative of this constant becomes zero so we could ignore it and the derivative of 4x^2 is 8x so now we can plug in point 1 comma 2
20:00 - 20:30 2 * 3 is 6 * 1^ 2 * 2^ 2 and we're going to have 8 * 1 so 2^ 2 is 4 * 6 that's 24 + 8 so that's 32 so that's the value of the first derivative with respect to X at the point 1 comma 2 now let's do the same for FY so first we got to find it let's treat X as a constant the derivative of y 2 is 2 Y and the
20:30 - 21:00 derivative of y Cub is 3 y^2 and 4x^2 is a constant so that becomes zero now let's plug into 0.1 two 5 * 3 is 15 so here we have 2 * 2 * 2 which is 8 2^ 2 is 4 * 15 that's 60 8 + 60 is 68 and so that's all you need to do in
21:00 - 21:30 order to evaluate the derivative at the point find the derivative and plug in the X and Y values here's another example but this one might be more trickier let's say we have a function it's equal to X Cub e raised to 4X y^2 and let's evaluate the partial derivative with respect to X and Y at the point 2 comma 0 so how can we find FX and
21:30 - 22:00 FY now here's we need to be careful when looking for f ofx you need to use the product rule because the X variable is in the first part of the function and it's in the second part but when finding FY you don't need to use the product rule so for example let's say if you have 4 x 5th you have a constant multiplied by a variable you do not need to use the product rule here this is simply going to be just 20 x to
22:00 - 22:30 4th now let's say if you have 4x^2 time e the 3 here you have a variable and a constant you don't need to use the product rule this is simply going to be 8X e 3r however if you have 4x^2 * e^ 3x you have a variable and a variable here so you do need to use the product rule so in the case for this problem
22:30 - 23:00 when differentiating with respect to X we have an X variable here and here so we got to use the product rule but when differentiating with respect to Y X is treated as a constant so we have a y variable on the right but we don't have a y variable on the left X cube is straight as a constant so we don't need to use the product rule in that case so I want to just help you to see when you should use the product rule and when you shouldn't so let's start with FY because we don't
23:00 - 23:30 need to use the product rule so it's going to be the constant which is X cub and then all we got to do is differentiate e to 4X y^2 which is going to be the same thing e to 4X y^2 times the derivative of 4X y^2 so 4X is a constant and the derivative of y^2 is 2 y so now we can plug in a point 2 comma 0 so X is 2 and this is going to be 2 to the 3 and we're going to have e raised to 4
23:30 - 24:00 * 2 * 0^ 2 0 * anything is 0 and then it's going to be 4 * X which is 2 * 2 and Y is z so because we have a zero here the whole thing is zero so that's the partial derivative with respect to Y now let's find it with respect to X so let's review the product rule let's say if we have two functions f * G you
24:00 - 24:30 need to differentiate the first function so you're going to get F Prime Time the original of the second plus the original of the first times the derivative of the second so that's how we're going to use the product rule so first let's differentiate the first part X Cub which is 3x^2 and then we're going to rewrite the second part we're not going to change it plus we're going to rewrite the first part X cub and then differentiate the second part so the derivative of e to 4X
24:30 - 25:00 y^2 with respect to X is going to be e 4X y^2 and the derivative of 4X y^2 with respect to X is going to be 4 * y^2 where y square is a constant so because the variable X was in both parts we have to use the product rule now let's plug in point 2 comma 0 so X is 2
25:00 - 25:30 and because we have a y^ 2 4X y^2 will become zero so this will be e to zero again and we have a y^ squ here so this whole thing will be zero 2^2 is 4 and 3 * 4 is 12 anything raised to the zero power is 1 2 to the 3r is 8 and 4 * 0 is 0 so this is going to be 12 and 8 * 0 is 0 so the final answer is 12 so the first derivative with respect
25:30 - 26:00 to X is 12 and the first derivative with respect to Y is zero consider this problem let's say F ofx comma Y is equal to 3 y to the 5th / XB + YB so go ahead and find a partial derivative the first one with respect to X and with respect to y now here's a question for
26:00 - 26:30 you do we need to use the quoti rule if you recall let's say if we wish to differentiate F / G where f and g are both functions that contain variables it's going to be GF Prime minus FG Prime over G ^2 now when differentiating with respect to X 3 y 5th is a
26:30 - 27:00 constant and X Cub is a variable so we don't have a variable on the top and on the bottom so for f ofx we don't need to use the quotient Rule now when differentiating with respect to Y we have a y variable on the top and on the bottom so we do need to use the quotient rule with respect to Y and so that's how you can tell so let's start with f ofx where we don't need to use the quoti rule
27:00 - 27:30 so imagine if you were differentiating let's say 4 over X Cub + 7 if that's the case what you could do is rewrite the function as 4 x + 7us 1 and simply use the power rule that's what we're going to do in this example so first I'm going to rewrite it as 3 y 5 * X Cub + YB raus
27:30 - 28:00 one so f ofx is going to be the constant 3 y 5 then I'm going to take the negative 1 move it to the front so it's going to be times1 and then I'm going to keep the stuff that's inside the brackets the same subtract 1 by one and that's going to be NE -2 and then differentiate the stuff on the inside the derivative of x Cub is 3x2 and the derivative of y is z because
28:00 - 28:30 we're going to treat it as a constant so this is going to go to the bottom making this exponent positive and everything else will stay on top so we got 3 * 3 which is 9 plus a negative sign so it's -9 x^2 y 5th / by XB + YB 2 so that's the first first derivative with respect to
28:30 - 29:00 X now let's find FY the first derivative with respect to Y and we're going to have to use the quo room so f is going to be the top function 3 y 5th and G is the bottom part X Cub +
29:00 - 29:30 YB now F Prime with respect to Y it's going to be 3 * 5 y 4 which is 15 y 4 G Prime with respect to Y is going to be 3 y^2 the derivative of x Cub which we're going to treat as a constant is zero so now let's plug it in into this formula here so F of Y is going to be G which we know G is just X Cub y
29:30 - 30:00 Cub time F Prime which is 15 y 4 minus F which is 3 y 5 time G Prime which is 3 y^ 2 all divid G ^2 which is X Cub + y^ 2 so now let's see what we can do to
30:00 - 30:30 simplify this expression so if we want to we could distribute 15 y 4th or we can Factor Out 3 y 4th let's Factor Out 3 y 4th if we do so we'll be left with a five times XB + YB so that's 5
30:30 - 31:00 XB + 5 YB now here we took out the three and Y 4th so we're left over with Y and that Y is going to be multiplied by 3 y^2 so it's 3 YB don't forget about the negative sign divided by this stuff so we can combine 5 y cub and 3 YB so
31:00 - 31:30 the final answer is 3 y 4th * 5x Cub + 2 YB / X Cub + y^ 2 so that is f of Y now let's say if you're given a function that looks like
31:30 - 32:00 this x^2 y cub and you want to find the slope of the surface in the X and Y Direction at the point 32 all you need to do is find FX and FY and evaluate the function at this point and that's going to give you the slope of the surface in the X and Y Direction so to find the slope in the X Direction it's simply FX 3 comma
32:00 - 32:30 2 so we just got to plug in well first we got to find FX FX is going to be the derivative of x^2 which is 2x time the constant y Cub so FX 3 comma 2 is going to be 2 * 3 * 2 the 3 power 2 * 3 is 6 2 the 3 power is 8 and 6 * 8 is 48 now let's find
32:30 - 33:00 FY that's going to be x^2 * 3 y^2 and so FY the slope of the surface in the y direction is going to be this is FY 32 it's going to be 3^ 2 * 3 * 2^ 2 3 2 is 9 and 2 2 is 4 3 * 9 is 27 and 2 7 * 4 is 108 so these are the answers that we're
33:00 - 33:30 looking for now sometimes you may have a function with three variables instead of two let's say we have f ofx y and z and is equal to X 5th y^2 Z to 4th now let's find the first three partial derivatives with respect to x y and z so what is the partial derivative with
33:30 - 34:00 respect to X now the rules are the same we're going to treat X as the variable and every other variable we're going to treat it as a constant so the derivative of x 5 is 5 x 4th so y^2 and Z 4 will be treated as a constant and according to the constant multiple rule we just multiply the derivative of the variable times the constant to get the derivative of the function with respect to that variable so that's F ofx now let's find
34:00 - 34:30 F of Y so X the 5ifth is a constant all we need to do is differentiate y^2 which is 2 Y and Z to the 4th is the same so when looking for f ofx differentiate just x to the 5th when look at for f of Y differentiate only the y^2 part and we'll looking for f of Z differentiate just the Z to the fourth part so F of Z
34:30 - 35:00 is going to be X 5th y^2 * the derivative of Z to 4th which is 4 Z cub and so that's how you can find a partial derivative with respect to X Y and Z so let's say w is equal to Z the 3times Ln zy time e raised to the
35:00 - 35:30 x^2 y Cub Z 4 so find DW DZ DW Dy and dwdx feel free to pause the video and be careful with this one so let's start with DW DX so do we need to use a product rule for this in terms of X we have a constant and a constant and a variable so we do not need to use the product
35:30 - 36:00 rule so Z 3 is a constant so we can rewrite it Ln zy will be treated as a constant and the derivative of e to the U is e to U * U Prime so it's going to be the same thing e x^2 y Cub Z 4 time U prime or the derivative of X2 y Cub Z 4 X2 is the only variable so we just got to differentiate that part which is 2X X time the constant YB Z 4 so that's
36:00 - 36:30 dwdx now what about DW Dy do we need to use the product rule it turns out we do here we have a constant here is a variable because there's a y in it and here this is another variable so we got two variables multiply to each other so therefore we need to use this form of the product rule the derivative of two variables mult mply to each other is derivative of the first * the second plus the first
36:30 - 37:00 time the derivative of the second so the first part is this part because that's the first time the variable shows up and this is the second part and that's just a constant so we can rewrite the constant Z the 3 and then let's differentiate the first part Ln zy the derivative of Ln U is U Prime over U so the derivative of zy is simply Z * 1
37:00 - 37:30 or Z we'll just differentiating the Y variable divided by the original zy times the other function which is e x^2 y Cub Z to 4th so this part is like f Prime and this is G now for the second part so let's rewrite the constant Z the 3 f
37:30 - 38:00 is basically the first time we see the variable Ln zy which we're just going to rewrite it as Ln zy and then G Prime the derivative of the second part that's going to be e to the U which is e x^2 y Cub Z 4 and then times U Prime which we're differentiating the Y variable so it's going to be X time the derivative of YB which is 3 y^2
38:00 - 38:30 * the constant Z 4th and I'm running out of space so let me get rid of this formula and let's simplify what we have so we can cancel a z and we can combine Z to the 3 and Z to the 4th so dy I mean DW Dy is equal to to Z 3r time e x^2 y Cub Z 4 / y
38:30 - 39:00 plus 3x^2 y^ 2 Z the 3r * Z 4th that's Z to the 7th and then Ln zy * e x^2 y Cub Z 4 if you want to you can factor out e x^2
39:00 - 39:30 y Cub Z 4 and you can also take out Z to the 3 if you want to but we're not going to worry about that so I'm just going to leave the answer the way it is now let's find the other one DW DZ so notice that Z is found in each term so we need to use a product rule with three variables
39:30 - 40:00 that's f * G * H so it's going to be the derivative of the first part F Prime * the other 2 plus F time the derivative of the second part * h plus FG * the derivative of the third part H Prime so in this particular example Z the 3 is basically F Ln z y Y is G and H is the E raised to all of
40:00 - 40:30 that stuff so let's find F Prime that's going to be 3 z^ 2 times everything else Ln zy and e to the x^2 y Cub Z to 4th plus F which is z 3r time G Prime so that's going to be U Prime over U the derivative of zy with respect to Z
40:30 - 41:00 is simply y divided by the stuff on the inside which is z y and then times e to the x^2 y Cub Z 4 and then plus the third part which is going to be Z the 3 time Ln zy times the derivative of this thing which is e x^2 y Cub Z 4 time the derivative of the exponent
41:00 - 41:30 which is x^2 YB * 4 Z3 so that's the answer for DW DZ you have to use the product rule with three different variables when I mean three variables I mean FG and H but Z is just one variable that's the only thing that we differentiating now let's see if we could simplify our expression let's factor out the greatest common factor so we can take out e to
41:30 - 42:00 the x^2 y Cub Z to 4th that's a common term and we can also cancel a y and z will cancel so that becomes a z^2 so we could take out a z^ s from every term so let's put that here now so this is gone and we took out the
42:00 - 42:30 z^ S so we're left over with three * Ln zy and the second term we took out this all of that is gone and we took out all of the Z squ so there's nothing left in the second term so we could just put a one for the last last term we took this out we took out z^2 so we're left with z
42:30 - 43:00 and we have Ln zy and everything here so it's going to be 4 x^2 y cub and then Z 3 * Z that's going to be Z to the 4th and then Ln zy so we can write the final answer like that if we want to now let's say if we have this
43:00 - 43:30 function and all we want to do is evaluate the first partial derivative with respect to Y at the point 1 comma 2 comma 3 so X is 1 Y is 2 Z is 3 go ahead and try that so first let's find FY so the derivative of y^2 is 2 Y the
43:30 - 44:00 second term doesn't have a y variable so the whole thing is a constant that becomes zero and the derivative of y is one so that's FY so now we can evaluate it at the point 1 2 3 so X is 1 and Y is 2 and z z is three so 2 * 2 is 4 and 3 to the 4th power is
44:00 - 44:30 81 that's 3 * 3 * 3 * 3 4 * 4 is 16 now 5 * 8 is 40 so 5 * 80 is 400 and 5 * 1 is 5 so 5 * 81 is 405 + 16 that's 421 so that's the rate of change of the function with respect to y at the point 1 2 3 so sometimes you may be asked to
44:30 - 45:00 identify just one of the partial derivatives at a certain point and that's how you do it now sometimes you may need to find Hower order partial derivatives so let's talk about the difference between the first derivative and the second and a different forms in which they can come in so let's say we have our original function f if we differentiate f with respect to X
45:00 - 45:30 we can get the first derivative with respect to X which is known as F ofx we could also differentiate f with respect to Y and we'll get the first partial derivative with respect to Y which is known as FY now at this point we could differentiate the first derivative with respect to X so the second derivative is going to be F XX which means that we found the derivative with respect to X twice now
45:30 - 46:00 we could differentiate FX with respect to Y and we'll get fxy which means that we differentiate it first with respect to X and then second we differentiate it with respect to Y now FY we could differentiate it with respect to X and we'll get fyx which means that we differentiate in first with respect to y and then with respect to X or we could differentiate
46:00 - 46:30 FY with respect to Y which is FY Y which means we differentiate the function twice with respect to Y so when dealing with second partial derivatives if you have a function with two variables there are four derivatives that you can get now let's say if we have the function x 3r + 4x X 5th y Cub + 5 y
46:30 - 47:00 4 so find all four second partial derivatives so find FXX fxy fyx and FY y feel free to pause the video so first let's find FX let's find the first partial derivative with respect to X the derivative of x Cub is
47:00 - 47:30 3x2 and the derivative of x 5th is 5 x 4th times a constant y cub and 5 y 4 is a constant so that's going to be zero so that's FX which we can write as 3x^2 + 20 x 4 y Cub so now let's find F double X so 3x^2 will become
47:30 - 48:00 6X and x 4 is 4X cub and 20 * 4 is 80 so we can write it as 80 x Cub y Cub so that's our first answer now let's find F XY so let's differentiate this function with respect to Y 3x2 would become zero and the derivative of y to the 3r
48:00 - 48:30 is 3 y^2 so this is 60 x 4 y^2 and so that's it for this one now let's find FY so we could find fyx and FY y so X Cub is now a constant that becomes zero and Y Cub the derivative of that is
48:30 - 49:00 3 y^2 and a derivative of y 4 is 4 YB so FY is equal to 12 x 5th y^ 2 + 20 y Cub now from this function we could find the other two answers we're looking for so let's find f y let's differentiate it with respect to Y one more time so it's going to be 12 X 5 * 2 y + 20 * 3 y^
49:00 - 49:30 2 so that's equal to 24 x 5th y + 60 y^2 so that's FY now let's find fyx so let's differentiate this function with respect to X the derivative of x to 5th is 5 x to 4th and with respect to X Y Cub is a
49:30 - 50:00 constant so 20 y Cub becomes zero so the final answer is 12 * 5 which is 60 x 4 y^2 so that is the second derivative where we differentiate with respect to Y and then with respect to X here's another problem let's say f is equal to X Cub y 4th plus
50:00 - 50:30 z^2 X Cub y + 4 x Cub Z to 4th so this time we're going to find a mixed partial derivative but a third derivative instead of a second order derivative so let's find the mixed third order derivative let's say F XYZ go ahead and find it so first we got to find F ofx we got to differentiate the function with respect to X so X Cub
50:30 - 51:00 will become 3x^2 and then we have another X Cub which is 3x^2 again * Y and the derivative of 4X Cub is 12 x^2 Z 4th so now we're going to differentiate the function with respect to Y so y to 4th will become 4 y cub Cube and the derivative of y is
51:00 - 51:30 1 now 12x^2 Z 4 doesn't have a y variable so that becomes zero now we're going to differentiate fxy with respect to Z now this term doesn't have a z so that's going to become zero z^2 is 2 Z * 3x^2 so the final answer is 6 ZX2 or 6 X2 Z now let's try another example using the
51:30 - 52:00 same function this time let's find f y x x so first let's find FY so this will become zero because there's no y variable the derivative of y^ 4 is 4 y cub and the derivative of y is is simply one so this is what we have left over now let's find fyx let's
52:00 - 52:30 differentiate this function with respect to X the derivative of x Cub is 3x^2 now let's differentiate that function again with respect to x 3 * 4 is 12 and the derivative of x^2 is 2X and the derivative of 3x^2 is 6X so the
52:30 - 53:00 final answer is 24 x y Cub + 6 x z^2 so that's the third order mixed partial derivative where we differentiate y first and then X and then by X again there's a theorem that you need to know related to the section and that is the equality of mixed partial
53:00 - 53:30 derivatives so if the function is continuous then you need to know that fxy is equal to fyx so if you differentiate with respect to Y once and with respect to X once the order doesn't matter these two mixed partial derivatives are equivalent to each other so let's say if you want to find a third order derivative f of XY Z is the same as F of
53:30 - 54:00 Y XZ which is the same as F of Z XY all of these three mixed partial derivatives are the same we differentiated with respect to X once with respect to Y once and with respect to Z once the order doesn't matter so here's another example F of Y XY is equal to f of x y y which is equal to F of Y YX these are all the same we're
54:00 - 54:30 differentiating with respect to X once but with respect to Y twice so let's prove it so let's say we have a function that's x^2 y Cub + 4x y^2 and let's prove that f ofx y is equal to fyx so let's find F ofx
54:30 - 55:00 first f ofx is 2x y Cub + 4 y^2 now let's find FX so let's differentiate it with respect to well let's find f of x y i at the derivative of y Cub is 3 y^2 * 2x that's 6X y^2 and the derivative of 4 y^2 is 8 y so now let's find fyx so first we got to find FY using
55:00 - 55:30 this function the derivative of y Cub is 3 y^2 so we're going to have 3x^2 y^2 and the derivative of y^ 2 is 2 y * 4X that's 8X y now let's differentiate this function with respect to X the derivative of 3x^2 is 6X * y^2 the derivative of 8 8X is simply 8 so notice that these two are the
55:30 - 56:00 same fxy is equivalent to fyx now let's say that f is 4 x^2 y Cub Z + 6 x y^ 2 z^2 and let's show that f ofx y y is equal to F of
56:00 - 56:30 yxy and let's show that's equal to f y YX so first let's find FX let's differentiate this function with respect to X so the derivative of 4x^2 is 8x times the constants y Cub Z the derivative of 6X is 6 and then time y^2 z^2 now we need to differentiate FX with respect to Y so it's going to be the derivative of
56:30 - 57:00 y Cub which is 3 y^ 2 * 8 so that's 24x y^ 2 Z and a derivative of 6 y^2 is 12 y * z^2 now let's differentiate it again with respect to Y so y^2 becomes 2 y * 24 so that's 48 XYZ and a derivative of 12 y Y is just 12 so let me write this
57:00 - 57:30 somewhere let's just separate this so f ofx y y is equal to 48 XYZ + 12 Z2 let's just write that for a reference we'll compare it later so now let's find f y y XY so first let's find
57:30 - 58:00 FY the derivative of y Cub is 3 y^2 * 4 so that's 12 x^2 y^2 Z and the derivative of y^ 2 is 2 y * 6 which is 12 XY Z ^ 2 now let's differentiate that function with respect to X the derivative of x^2 is 2X so 12 x^2 is 24x * y^2
58:00 - 58:30 Z and a derivative of 12x is just 12 y z^2 now let's differentiate that function this one with respect to Y so the derivative of y^2 is 2 y * 24 so that's 48 XYZ plus the derivative of 12 Y is 12 so notice that these two are the same now let's find the last
58:30 - 59:00 one f y YX so let's find FY first which we know it's 3 y^2 * 4 so that's 12 x^2 Z and y^2 is 2 y * 6 that's 12 XY z^2 now let's find the derivative of this function with respect to
59:00 - 59:30 Y and it should be a y s here which I almost forgot to write so the derivative of y^2 is 2 y * 12 that's 24 x^2 y z and the derivative of y is 1 so times everything else which is going to be 12x z^2 now let's differentiate this function with respect to X the derivative of X2 or 24x ^2 is going to be 48x and a derivative of 12x is just 12
59:30 - 60:00 so as you can see we're going to get the same thing so the order doesn't matter the reason why these are the same is because we're differentiating with respect to Y twice and we're only differentiating with respect to X once but the order doesn't matter so mixed partial derivatives are the same if you differentiate each variable the same number of times regardless of their
60:00 - 60:30 order so that's it for this video so hopefully you have a good idea of partial derivatives and now you know how to find them so thanks for watching