find-the-partial-derivative-of-the-dependent-variable-or-function-with-respect-to-each-of-the-ind-59

Question

Find the partial derivative of the dependent variable or function with respect to each of the independent variables.
$$z=\left(3 x^{2}+x y^{3}\right)^{4}$$

Tyler Moulton · Accepted Answer

Okay we want to find the first partial derivatives of the function I have X. Y. Is he going three X squared plus x Y cube to the power of four. This question is challenging our understanding of differentiation and applied to our newfound multi variant functions. So the first partial divers are searching for our DF DX and DY where for each we have to use standard differentiation techniques. We learned in single very real calculus. But for DF dx, n E Y term street is a constant and for the FBI any extreme street is a constant. So for fx we have to start off with a chain role. So a derivative ist four times three X squared plus X Y cube cubes 10 6 X plus Y cube. Where six X Y cube is different is a derivative of what&#x27;s inside parentheses. Similarly, F Y is four times three X squared plus X Y cube cube time, zero plus three X Y squared. The reason there is now a zero is because the three X squared and fantasies does not have a term of why they&#x27;re what&#x27;s constantly goes to zero. Thus fy is 12 X Y squared term three X squared plus x Y squared, cute.

Linh Vu · Answer

and, of course, moving. Actually she Thanks quiet. See in cartoons You x square one minus a tree. Ex&#x27;s email for Brust 10 y Square Z Square Now the first pressure area with the words went to the X So X will be the variable, the others on the constant. So it&#x27;s again for thanks, boy. Minus through an a minus three z bo for and now a brushing her If there was rented a why we get no one with a variable, so have a two x square and then plus 21 say about two. The bus to the roof of yours. But you dizzy now a single bit of arable so have done minus trial xz about three and then we have the blessed 20 wise squares e

Linh Vu · Answer

have you X y c e w square x who I square for us. Thanks. Why about trees? A square knot? You find a passion there. If there was what you did, step on you so that you could be there Variable did not That&#x27;s will be there and so have a chilled up a u x y square. And now the partially rift was watching Next extra with a variable, the orders will be a constant. So how do you square? Why square? Because why about three The square especially rift? And was that why? Why with a variable, the others as constant. So getting a call agenda to w square x qy three Thanks y squares the square And the last one brush of the rift It was pretty to see Gonna go Geo Now ju Thanks 13 c.

Bobby Barnes · Answer

So if we want to find this mixed partial fraction here, the first thing we&#x27;re going to do is just start taking the derivative. So over here, we can kind of look and see. Okay, Well, here we need to take three with respect to see, because it&#x27;s Dell by execute. Over here, we have dealt by. Why Square? So that&#x27;s going to be, too. With respect to why. And then we just have Delic. So that&#x27;s just gonna be one derivative with respect tax. And so, technically, we would start on the right side and then work our way toe left. But, um, in most cases, it&#x27;s really not gonna matter The order we do this in, but just kind of for sake of notation that they have here, we&#x27;re going to go from right to left. So over here, if we were to do so, Dell, by del Z of this, this is going to give. So we have Dell, you by Del Z. Um, now we&#x27;re going to treat and actually just write this out this first time, and then maybe the next couple of times I won&#x27;t do this. Now remember what we&#x27;re going to do is treat X and y here as Constance. So since we&#x27;re taking the derivative of something and we have a constant being multiplied by it, we just go ahead and factor that out. So just like we did in a single variable case, we can factor those out. So we have extra a light to the B and then del by Del Z of Ecstasy. And now we can just use power rule to take the derivative of this. So remember, Powerful says we take the power, move it out front, subtract one off of it. Um so that is going to leave us with X to the A. Why to the b. And then this would be Z actually see times Z rays, the C minus one. And so this is going to be tell you, not d, but del you by Del Z. So we have our first partial. Now we can go ahead and do this again with respect to Z once more And so again we treat these Tua&#x27;s constant so we can pull them out front so we would have the following. So we have elsewhere. You by del Z squared it is going to be equal to X the A Why? To the b And actually let&#x27;s take this C and move it out front. So it be time. See, You know we have Dell by del Z of Z two of C minus one. And then again we use power rule for this. So this would be C and minus one times Z to the C minus two. And if we were to write that out So we have Dele squared you over, Del Z squared is equal to see time C minus one next to the A why to the beat Z to the C minus two. And since we need a third partial of this with respect to Z, we go ahead and apply this one more time. Hotel Baia del Z on each side that more like a d as opposed to it too. That&#x27;s gonna give So now, Dell Cube of you and then D by Deasy cute and once again, all of this out front we&#x27;re treating is a constant. So we just pull it out So it be see Time&#x27;s see and by this one x today y to the B del by del Z of mad equals, but all Z to the C minus two. And now this. Here we take the partial is going to be C minus two times e to the C minus three, and I Let&#x27;s just go ahead and move that constant out front. And that would give So the third partial derivative of you with respect to Z is see time&#x27;s C minus one time C minus two times x to the A y to the B Z to the C minus two. All right, now, um, we need to take two partials with respect to why, If I&#x27;m not mistaken, yes, because we have del y squared, so we need to do that twice. So let&#x27;s come up here and do that. So you have dill Beidle. Why? And we apply this to each side. So over here, we&#x27;re gonna have dealt to the fourth you over now, in the denominator since, um, we have del y and l z, we just kind of keep those in the denominator kind of by themselves. So del y tells cute right now over here on the right side. So we&#x27;re taking the derivative with respect, toe. Why now? So that means So all these constants, I mean, they&#x27;re still Constants X is a constant and Z is a constant, so we can pull those out. So bc times c minus one times C minus two x The a z to C minus two and then del Beidle Why of why? To the beef? Yeah. Now we could go ahead and take the derivative of this using power rule. So be times y to the B minus one. And then let&#x27;s go ahead and pull that constant out front. So it be B times C C minus one C minus two and then x to the A. Uh, why do the B minus one and then Z to the C minus two? And so this is Dell to the fourth u D Y d z acute. Now we can go ahead and apply the partial respect. Why one more time. And so we&#x27;re trading. All of these is constants. The X is the constant and disease a constant again. So we just pulled those out when we take the derivative. So this would be dealt to the fifth you. And then, since we have another Dell, why we would have del y squared. So, Dell, why, uh, squared and then DZ cube on over here. We pull those out, so would be B C C minus one C minus two next to the A Z to the C minus two and then del Beidle. Why of why B minus one and again power rule. So it be B minus one times why to the B minus two. And then we&#x27;ll pull that constant out front. So BB times B minus one C times C minus one C minus two next to the a y to the B minus two z to the C minus two and then dealt to the fifth del Y squared is cute and so we&#x27;re almost done because now we just need to take the partial of this with respect. So we applied del Beidle X on each side, and now we treat all of these as constant. Since they&#x27;re already constant, we treat the y and the Z is constantly would go ahead and pull those out. And so, for the last part will have Dell six you over del Extell y square del Z cute. And so this would be times B minus one C c. minus one C minus two. Um, and then we have Y B minus two Z C minus two. And then Del by Dell X of X to the A. We use powerful for this, So move the power out front. So a times X to the A minus one and then we would be done. So let&#x27;s just go ahead and clean this up a little bit so we would get our final answer being so dealt to six. You over Del Extell. Why squared Del Z? Cute is equal to so a times B B minus one times. See, See by the swan seat by this, too. And then x to the A minus one wanted to the B minus two z to the C minus two. I mean screen a little bit. And this here is our final answer. And so again, you just you don&#x27;t necessarily need to do it. In this order here. You could have actually delight first brush with respect to X. Then why and then Z and you would have got the exact same answer because we can pretty much do these in any order. Um, since this is what we would consider a smooth function. But I mean, other than that just kind of following the notation, they have their though Either way, you would end up with the exact same partial here.

Problem

Find the partial derivative of the dependent vari…

Problem 15 Easy Difficulty

Find the partial derivative of the dependent variable or function with respect to each of the independent variables.
$$z=\left(3 x^{2}+x y^{3}\right)^{4}$$

Answer

$\frac{\partial z}{\partial y}=12 x y^{2}\left(3 x^{2}+x y^{3}\right)^{3}$

Related Courses

Basic Technical Mathematics with Calculus

Related Topics

Discussion

Top Calculus 3 Educators

Anna Marie V.

Samuel H.

Michael J.

Joseph L.

Calculus 3 Courses

Vectors Intro

Vector Basics - Example 1

Recommended Videos

Watch More Solved Questions in Chapter 29

Video Transcript

Allyn J. Washington, Richard S. Evans

Basic Technical Mathematics with Calculus

Anna Marie V.

Samuel H.

Michael J.

Joseph L.

Vectors Intro

Vector Basics - Example 1

What do you need help with?

Textbooks

Ask a question

Courses

AI Tutor

$\frac{\partial z}{\partial y}=12 x y^{2}\left(3 x^{2}+x y^{3}\right)^{3}$

Basic Technical Mathematics with Calculus

Anna Marie V.

Samuel H.

Michael J.

Joseph L.

Vectors Intro

Vector Basics - Example 1

Allyn J. Washington, Richard S. EvansBasic Technical Mathematics with Calculus

Anna Marie V.

Samuel H.

Michael J.

Joseph L.

Allyn J. Washington, Richard S. Evans

Basic Technical Mathematics with Calculus