use-implicit-differentiation-to-find-d-y-d-x-x2-yx-y26

Question

Use implicit differentiation to find $d y / d x$.
$x^{2} y+x y^{2}=6$

Maneesh Budda · Accepted Answer

s O today will be solving this problem. So this problem is X squared times Why plus X Y squared is equal to six. So when doing the implicit differentiation of this equation don&#x27;t forget to, uh, do the chain rule for each of these multiplication Sze. So first you want to take the derivative of X squared which will be two x times Why, plus X squared times too dirty of why Which will be one and you don&#x27;t want to forget uh, including d y over dx cause whenever you take the differentiation of and why variable you want to include ah d y over DX. This is what is known as implicit differentiation So plus the derivative of X which will be one times why square plus x times the dirt of the derivative of y squared Which will be to why times don&#x27;t forget to put d x again d y over DX. My bad is equal to zero because you have to take the derivative of six which will which is a constant and so become zero. So after that, you want a group all the d y over DX variables together and ah and the rest of the other components in this equation together, So this will turn out to be two x. Why plus y squared plus x squared times D y over d x close to x y comes D Y over The ex is equal to zero And now you want to take all these variables that don&#x27;t have to be Why, over the X component to the other side So you could single out this d y over dx um, in place so that you get the implicit differentiation of this equation. So it becomes that&#x27;s a square times D y over DX because to X y times d y over d x is equal to negative to x y minus y squared cause he&#x27;s a practice this on both sides So it will come out to negative to X uh, why I minus y squared and then you have to cept substitute this do whatever d x variable out of the equation. So we d y over d x times X square crush two x Why is equal to negative to X minus y squared. So if you take the distributive property and you do, um, you distributed to each of these terms, you&#x27;ll get X squared times do whatever the X plus two x y 10 31 of dxl, which is the original of equation back in this step. So now, um, once using a lot does d y over DX. You take this other other side by dividing on both sides, so turn out to be d y over d X is equal to negative to x y my nice way square over x squared cause to X. Why, when this is a solution for the implicit differentiation of dysfunction. I hope this helped, um, if you have any questions, just reach out to me and, uh, thank you.

Maneesh Budda · Answer

hi, guys said today. We&#x27;ll be taking that implicit differentiation of this equation. So whenever there&#x27;s multiplication between two variables, don&#x27;t forget to do change rule. And so let&#x27;s start off by taking the curative of two times x times. Why? So it&#x27;s too, and you want to keep the two outside cause it&#x27;s Ah, it&#x27;s a coefficient for the X and Y, very bored. And you only want to take the curative between the multiplication of these two variables. So we&#x27;ll keep that outside and we&#x27;ll put prentiss parentheses. And so we&#x27;ll take the differentiation of X, which is one times why, plus the differentiation of, uh, why, um Times X So will be just x times one times d y o r d x. So whenever you&#x27;re taking the derivative of the X or I mean the Y variable, don&#x27;t forget to include D y ver d x. This is what is known as implicit differentiation. And then we put this in there. We close it off with the parentheses so that this, too, is only applying to these two terms, and then we at and we take the differentiation of why squared, which is to why times do you? Why over the ex is equal to And don&#x27;t forget to take the derivative of X plus y, which is just one plus one times D y over DX. Remember that you have to take the you have to write. Do whatever the ex whenever you&#x27;re differentiate different shooting. Why? And so you want a group All of the D white over DX terms together and the rest of the other terms together. So two applies to If we distribute, too. Don&#x27;t become, too. Why cause to eggs Times D y over d. X plus two. Why times D y over d X is equal to one plus. Do you? Why over d X Now we have to take this d Y o r d x on the right side, over to the left side, and you have to take two. Why over to the right side. So if we subtract to why on both sides will get to X times do y over d X plus two y times D y over D X is equal to one minus two. Why close do you wire over D X and now we have to subtract do whatever dx on both sides D y over the eggs and minus do you I over dx So this will turn out to be two x times d y over d X cross to Why times do you i over d X minus d y over D X is equal to one minus two. Why I&#x27;ll go ahead appear and continue selling the problem up here. So once we got all the do I ever d eggs terms group together, we wantto take the d y over dx term out so that we could isolate. Just do whatever d x from the equation. So we do. Whatever the x, Times two X plus two Why a number get minus one Cause you see this negative right here Don&#x27;t account for that is equal to one minus two Why and then after isolating the d y o r d x, you can see that once we distribute the this do I rdx into each of these terms, you&#x27;ll get the same equation as the previous step. So now we divide two x plus two y minus one on both sides. So if we do that, it will cancel out on the left side completely and will be moved to the right side of the equation. So then you&#x27;ll get D Y over D. Eggs is equal to one minus two. Why over two X plus two. Why minus one. I hope this helped and see you guys later.

Tom Greenwood · Answer

all right, We&#x27;re trying to find the derivative of this expression. You know, we have to use implicit differentiation because we cannot get this solved for why, all by itself. And we&#x27;re taking the derivative of lie with respect to X. I hope you don&#x27;t mind. I&#x27;m not going to use that notation. What? Some working this out I&#x27;m going to use. Why, Prime? Just to save myself some writing and some space, you&#x27;re on my board. So the first thing you have to do is we have to look at our expression and remember the different rules. For starters, two x y that&#x27;s gonna use the product rule. We have function two x here times why I&#x27;m gonna treat it that, say, as I do my product group. So we would have the first function to x times the derivative of the second function. So the derivative a Why is r t y d x or y prime? Plus the second function? Why times the driven off the first in the derivative of two X is just too. And now we can move on to the rest of the expression. So plus derivative of y squared will be to Why? Using the power room Times the deal i d x again times y prime equals driven of X is one plus the derivative of Why is why prime? So now we need to look at this and we need to get all our white crimes together. So we&#x27;ve got two terms here with a Y prime and we&#x27;ve got one more over there. So I&#x27;m gonna get all three of those on the same side. And as I do that I&#x27;m gonna factor out the UAE prime from all three of those so that I&#x27;ll give me too X plus two. Why? And then when I moved the white prime over, I&#x27;m gonna be subtracting it, so that would be a minus one. Why prime? So just minus one. And then I factor out the UAE prime. So there&#x27;s all my crimes gathered together on one side, and then that means I need my other terms on the other side of the equation. So subtract the two. Why here from the one and we get one minus to lie on the other side of the equation. And now to get why, prime by itself, I just have to divide by I set of parentheses here on both sides. So when we divide both sides by two x plus two y minus one, we get why prime R D y d X that we were looking for equals the one minus two. Why divided by this whole set of parentheses of two X plus two I minus one. And there&#x27;s our derivative with respect X.

Tom Greenwood · Answer

all right, We&#x27;re trying to find the derivative of this equation, and we have to use implicit differentiation to do that. So taking the derivative of everything with respect to X, the derivative of Ito a function is e to the same function. But then, because of the chain rule, you have to multiply that by the derivative of that function. And here we have to use the product rule because we have two functions multiplied together. And so product rule is first function X squared times the derivative of the second function the driven of a Why is gonna be y prime or the d y d x. I&#x27;m going to use the prime notation and then plus the second function. Why times the derivative of the first function, the derivative of X squared is two X and that&#x27;s gonna equal the derivative of the right hand side derivative of two X is too. And then the derivative of two. Why will be two times or D y DX or the light prime again? Let&#x27;s do some distributing here to give it to these parentheses. We will have next square times y prime times e to the X squared y plus two x y times e to the x squared y and that illegal to pause to why prime? So now since I&#x27;m trying to solve this for why prime, I need to get my terms with the prime on the same side of the equation and my other terms on the other side of the equation. So just do some subtracting to move the two y prime over and to move that two x y e to the X squared y over to the other side. And when I bring my wife prime over because both of these have a wide prime, I&#x27;m gonna factor that out at the same time. So I&#x27;m going to have X squared times e to the X squared y minus two old times y prime. And that&#x27;s my left hand side. And the right hand side will be to minus the two x y he to the X squared y power. And now to get the UAE prime by itself, because I factored that out. So that would just be modification. And we can get rid of the parentheses just by dividing both sides by that entire set of parentheses. And so we end up with why prime equals tu minus two x Why? Okay to the X squared y all divided by what was in our parentheses. X squared e to the X squared. Why minus two And there&#x27;s our d y d X, or are wide prime whichever notation

Robert Daugherty · Answer

Okay, section 3.7, and just double check that real quick. So 3.7 implicit differentiation. Problem number five We&#x27;re giving this function and we are asked to differentiate this with respect to X. So I need to find what is the value off soon? Erase that. What is the value of d d X of this entire expression? So differentiating everything with respect to X. So first up, let&#x27;s look at the left hand side. I&#x27;ve got a product X squared times x minus y squared. So if we differentiate X squared, that&#x27;s where I get to X. And then that will be multiplied by X minus y squared. Plus right down the X squared. And now I need to differentiate two times x minus y Well, that is gonna be to X minus. Why? Okay. Times the derivative of what&#x27;s inside there. So that&#x27;s gonna be one minus. Do you? I d X, that&#x27;s the left hand side equals when a difference your right hand side. I get to X minus two. Why do you Why D x now that&#x27;s the end of the calculus. This all is algebra from here on out to try to simplify this because when I need to solve for is d Y d X. So let&#x27;s expand everything on the left. So I&#x27;m going to have to x times x squared minus two x y plus y squared plus Ah, it looks like I&#x27;ve got two x squared when I multiply X minus Y time one minus D Y t x, I get X minus. Why? And again minus x Do you I the ex? Plus why do you Why D X equal two x minus two I do. You? Why dx okay, And then there&#x27;s a parentheses that right there. Okay, so my goal get to you. I need to get d y t X all by itself. So let&#x27;s continue expanding. And so the next line I&#x27;m going to have what? To x cubed minus four x squared. Why? Plus two X Why squared and then I get plus two x cubed minus two X squared. Why minus two x cube D Y the X plus two X squared. Why D Y the X equal to X minus two. Why do you why d x Okay, so let&#x27;s see what can combine here. So it looks like I can have. Um I think what I would do is to put, um if you look at two x cubed and two x cube. So that combines to be four x cubed. It looks like the X squared y terms right here. So those will combine to be minus six x squared. Why? And then it looks like I&#x27;ve got plus two x y squared. So I&#x27;ve got that term taking care off. Okay. And then what I&#x27;m gonna do is write minus two ex. Bring that to the left side. Okay? And then everything with a do I d x. I&#x27;m gonna move over to the right side of this equation, so that&#x27;s gonna be equal to two X cubed minus two x squared Y minus two. Why all of that? Do you? Why d x? Okay, so I took care of everything. So for X cube minus six x squared y plus two x y squared minus two x equal to X cube monies to X squared. Why? Minus two? Why all of that? D y t x now. I wrote down all of that, but I&#x27;m run out of space and we need to go to the next line. So let&#x27;s just rewrite where we were. So we were at four for X cubed minus six X squared Y plus two x y squared minus two acts equal to X cubed minus two X squared. Why minus two? I do. You? Why the ex? So now I know that, Do you? Why d x is equal to four x cubed minus six X squared Why plus two x y squared minus two x All of that divided by two X cube monies to X squared y minus two. Why? And then what I can do now is I could do some factory ization. So if I factor um ah to out. So when we get ISS, do you? Why The X is equal to two two x cubed minus six x squared y plus X y squared minus X all of that over too. X cube minus X squared y minus. Why, to over two is equal to one. So my final answer is two x cubed minus six X squared y plus X y squared minus X all of that over X cubed minus X squared Y minus. Why? Okay, so there was a lot guys So if I look back at this question again, this is one of the big use cases off when people struggle with calculus. It&#x27;s not the calculus part. It&#x27;s the algebra. So if you get it so if after your first you know few months with algebra, this is just taking the derivative using chain roll product rule, that part was pretty straightforward. It was all the algebra that followed to solve her D Y D X, where it was real picky and mistakes could have been made.

Tom Greenwood · Answer

what? We&#x27;re trying to find the derivative of this expression and see equation, and we want to find d y d X, and we have to do that with implicit differentiation. And as I look at this thing, this left hand side, you&#x27;ve got the two things multiplied together and trying to do that with the product rule I think is going to be about more cumbersome. And if we do the algebra to simplify this so I&#x27;m going to square my binomial So that will give me X squared minus two x y plus y squared from just the rule about square in a binomial from outside. And then I&#x27;m gonna multiply that by the X squared and distribute that through. And so that means the left hand side will become X to the fourth minus two x cubed times Why plus X squared y squared and that&#x27;ll equal X squared minus y squared. So now I still have the product rule with both of these expressions here, but I don&#x27;t think that&#x27;s going to be as bad to work out as if we were doing the product rule with the original stuff. So depends on how you want to handle this. So not doing my derivatives driven about X to the fourth is gonna be four x to the third during the next one. I&#x27;ve got my minus to execute. Why? That needs to use the product rule. I&#x27;m going to use the minus two X cubed as my first function. And why is the second so? The product rule says, First function negative two x cubed times the derivative of the second, the derivative A Why is R D Y d X, or I&#x27;m going to use the white private protection Plus the second Why times the driven off the first, the derivative of negative two X cube would be negative six x squared. So there&#x27;s that one. And now for the next part, the X squared y squared that also uses the product rule. So same idea. First function Next squared times the derivative of the second. Which would be to why Times y prime plus the second function. Why squared times the derivative of first the derivative of X squared is two X and that&#x27;s gonna equal now the right inside. So the derivative of X squared is just two x minus the derivative of y squared is gonna be to Why times? Why Prime Again? Yes, These congee get pretty long expressions. So now we want to collect all of our terms that have a Y prime with them on one side and everything else on the other side of the equation. And I also want to factor out the wide prime from everything on that side of the equation. So everything with a y prime well, that expression is a wife crime. That expression has a white prime and then expression as a white prime. So when I move the two y y prime over, I&#x27;m gonna be adding that to both sides. And again, I&#x27;m gonna fact throughout the UAE prime from each of those terms. So that will give me negative two x cubed plus two x squared why? And then moving this to the other side, that becomes a plus two. Why those are all multiply times a white prime which, like I said, I&#x27;m factoring that out and then that all equal all the other terms. But again, I have to move them over to the other side of the equation. So the four x cubed has to get moved over so that would become a minus four x cubed the minus six x squared. Why get moved overs that would become a plus six x squared y The two x y squared is getting moved over to the other sides that have become a minus two x. Why square? And then finally, the two X it was already over here just stays plus two x. So now, to get the UAE prime by itself, I have to divide everything by the negative. Two x cubed plus two x squared y plus two i on both sides of the equation. Negative two x cubed plus two X Square Live plus two Why? And so now the white primes by itself. So that&#x27;s my D Y d X equals. And this is not my final answer, because if you&#x27;ll notice all these coefficients that all divide by two so I can factor it two out of the numerator and the denominator, and I can simplify that, so that would become negative. Two x cubed plus three x squared Why minus x y squared plus X, all divided by in the two dividing out of the denominator Negative X cubed plus X squared. Why, plus why

Problem

Use implicit differentiation to find $d y / d x$.…

Problem 1 Easy Difficulty

Use implicit differentiation to find $d y / d x$.
$x^{2} y+x y^{2}=6$

Answer

$\frac{d y}{d x}=\frac{-2 x y-y^{2}}{x^{2}+2 x y}$

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$\frac{d y}{d x}=\frac{-2 x y-y^{2}}{x^{2}+2 x y}$

Thomas Calculus

Catherine R.

Anna Marie V.

Caleb E.

Samuel H.

Precalculus Review - Intro

Functions on the Real Line - Overview

George B. Thomas Jr.Thomas Calculus

Catherine R.

Anna Marie V.

Caleb E.

Samuel H.

George B. Thomas Jr.

Thomas Calculus