Download the App!

Get 24/7 study help with the Numerade app for iOS and Android! Enter your email for an invite.

Get the answer to your homework problem.

Try Numerade free for 7 days

Like

Report

Find the partial derivatives with respect to (a) $x,$ (b) $y$ and (c) $z$.$$f(x, y, z)=2 x^{2} y^{4} z^{3} e^{3 x^{4} y^{2} z^{2}}$$

(a) $\left(24 x^{5} y^{6} z^{5}+4 x y^{4} z^{3}\right) e^{3 x^{4} y^{2} z^{2}}$(b) $\left(12 x^{6} y^{5} z^{5}+8 x^{2} y^{3} z^{3}\right) e^{3 x^{4} y^{2} z^{2}}$(c) $\left(12 x^{6} y^{6} z^{4}+6 x^{2} y^{4} z^{2}\right) e^{3 x^{4} y^{2} z^{2}}$

Calculus 3

Chapter 6

An Introduction to Functions of Several Variables

Section 2

Partial Derivatives

Campbell University

Oregon State University

Harvey Mudd College

University of Michigan - Ann Arbor

Lectures

12:15

In calculus, partial derivatives are derivatives of a function with respect to one or more of its arguments, where the other arguments are treated as constants. Partial derivatives contrast with total derivatives, which are derivatives of the total function with respect to all of its arguments.

04:14

04:43

Find the partial derivativ…

04:22

04:04

05:00

03:41

04:00

Find the indicated partial…

03:25

03:31

06:48

03:07

Find all three partial der…

So if we want to find the partial derivatives of this, remember, what we're going to do is we assume that all other variables are constantly that we just take the derivative like we would, um, in the single variable case with these constants floating around. So let's come over here. So we do, del by Dell X start to find the partial respect X That's gonna be DLF by Dell X is equal twos. We normally would. Writers just f sub x. Now this Why? And this is e here, we assume, are constants. So just like how I can factor that two out when I'm taking this Too rude, I can factor those out as well so we can write this as to why? To the fourth z cute and then Del by Dell X of X squared E to the three x to the fourth y squared ZIS work. Now we have two functions that depend on angst, that we're multiplying together so we can go ahead and use product all to take this derivative. So this would be to y to the fourth Z Cube. Um so I'll just kind of write out what this will be B e 23 x to the fourth, twice where C squared and then Del by Dell, X of X squared, plus X squared times, the partial derivative with respect to X of E +23 acts to the fourth y squared Z sward. Now the partial derivative of X Word is just going to be two X and then over here. To take this, we're going to have to use changeable, so stay the same. First would be e to the three x to the fourth. Why Squared z squared? And then we have to do del by Dell's Z X of what's on the insides of three x to the fourth y squared z sward. And then again, remember, we're going to treat y and Z is if they're constant, so we can pull those out. And then we just take derivative of three X to the fourth, which would give us 12 execute. So, actually, if I should do this in the green, so it be 12 x cubed y squared C squared. Um, and now I'm gonna factor these terms out here, and then I just kind of leave that, and then we could see what else? We can factor after that. But so if we do that, we end up with two y to the fourth z cooped E to the three x to the fourth widespread cease word time. So we'd be left with two x there and an X Square times that would be 12 x toothy fifth, um, y squared c squared. And then I think the one thing we could factor at this point is an ex. So let's just go. Oh, we can actually factor out two x So let's go ahead and do that. So it be four x y to the fourth z Q e d to the three x to the fourth y squared C squared and then one plus six x to the fourth. Why squared C square? And I guess it's not necessarily needed to factor it. Um, but I mean, if you can kind of clean it up, make it look a little bit prettier. Uh, then that's normally advised. So that's our partial with respect to X now to get our partial introspect why it will be the exact same process. But now we're going to assume that X and Z are our Constance. So how do Del by Dell? Why on each side? So they'll give l f by dell wise to so f sub y. So this exodus see here we assume our constant so I can pull those out First would be to x squared z cute gel by del y of why? To the fourth e to the three x 2/4 wise where c squared and we could go ahead and, um, apply products will just like we did before. So let me go ahead and write that out. So product will will have this he first and we do, del by del y of y to the fourth. Plus, um, Then we just switch those we have y to the fourth and then del by del y e do the three x four y squared c squared. Okay. And then again taken to the partial of White to the fourth, which is give us or why, Cube, just using power rule and then over here, um, so we use change will once again, it would be e to the three x to the fourth. Why squared C squared, Then we have Dell by Dell. Why of three x to the fourth y squared z squared. And then again, remember this X and Z. We can kind of like pullouts or really just taking the derivative of y squared, which would give us to buy. So then this is going to become six x to the fourth y Z to the fourth and again, all factor out that e term there, and then we'll see what else we can factor out after that looks. So now, in this first time, we'd be left with four y cubed. And then over here, if I multiply that y to the fourth with that, that would be six x to the fourth. Why? To the fifth Z to the fourth? Um, and it looks like I can pull out a to y cube. So it's going to do that. So it be four x squared y cube cube E to the three X to the fourth y squared cease word. Um, and then here I should just be left with two plus and then three x to the fourth. And then that would be y squared C to the fourth. And then this is our partial with respect to why and then, lastly for a partial respect to ze. Same thing we did before. But now we just assume X and Y are are constants. So tell by del Z so x and y we can pull those out So we have DLF by Del Z is equal to f sub Z. It would be two x squared y to the fourth and then del by del Z of Z cubed E to the three x to the fourth y square C squared. And then again, we use product rule. Try to find the derivative of that So two x squared y to the fourth so e to the three x to the fourth y squared z squared then del by del z Z Cube Boston. Then we switch them around again. Uh huh. And then we just take the derivative like we did the last two times. So first used product rule over here that would give us three z squared over here chain rule. So that is going to be e to the three X to the fourth Weiss, where C squared. And then del by del Z of, uh reacts to the fourth y squared C square and then remember this three x before y squared we treat is a constant. So we're really just taking the derivative of Z squared. So I'll just be to Z. So this is gonna be six x to the fourth. Why squared Times E. And then once again, I'll just go ahead and pull out that power of E r exponential and then we'll see what we have left to continue to pull out. So if I pulled that out, we should just be left with three z squared here. And then if I multiply the CQ by this of that should be six um x to the fourth. Why squared? Um, see to the fourth and then looks like we could pull out of the three z squared. So that would give us six x squared y to the fourth z squared, then e to the three x to the fourth y squared c squared. And then that first time, we would just be left with one. And then plus two x to the fourth. Why squared? And then we pulled out the disease first. I'll just use this word, and then this is going to be our partial with respect to Z. And so again, remember, when we're taking these partials, all we're really doing is oh, training all of the other variables as if there were constant and imply all the derivative rules that we learned for the single variable calculus case.

View More Answers From This Book

Find Another Textbook

01:57

Find and classify, using the second partial derivative test, the critical po…

01:30

Use the method of Lagrange multipliers to optimize $f$ as indicated, subject…

01:32

Evaluate the given integral and check your answer.$$\int 3 d x$$

06:35

Evaluate the given integral.$$\int x y \sqrt{1-x^{2}-y^{2}} d y$$

01:01

Find $f(x)$.$$f^{\prime}(x)=3 e^{x}-\frac{2}{x^{2}}+1, f(2)=7$$

02:11

Evaluate the given integral.$$\int\left(2 x^{2}-3 x y^{3}+y^{2}+3\right)…

Find the partial derivatives with respect to (a) $x,$ (b) $y$ and (c) $z$.

01:56

Evaluate the given integral and check your answer.$$\int\left(2 r^{3}-\f…

Find $f(x)$.$$f^{\prime}(x)=2+5 / x, f(1)=3$$

03:33

Find the partial derivatives with respect to (a) $x$ and (b) $y$.$$f(x, …