27–40. Implicit differentiation Use implicit differentiat...

Question

27–40. Implicit differentiation Use implicit differentiation to find dy/dx.

y = xe^y

Accepted Answer

Welcome back, everyone. In this problem, we want to find the derivative of Y with respect to X by implicit differentiation for Y equals x2e the negative Y. A says it's -2 X multiplied by X2 + E to the Y. B2X multiplied by X2 + E to the Y. C. 2 x divided by X2 minus E to the Y. And the D2 x divided by X2 + E to the Y. Now if we're gonna implicitly differentiate Y here, of note is that Y equals X2 E to the negative Y, a product. So we can differentiate our left hand side with respect to X with implicit differentiation and differentiate the right hand side again, still implicit differentiation, but using the product rule. So let's remind ourselves of the product rule before we continue, OK? Now recall. That if we have a term y equals UV where U and V are functions of X, then the derivative of Y with respect to X equals the derivative of u multiplied by V plus u multiplied by the derivative of V. In other words, we differentiate the function multiplied by the opposite original function and then add our results. Now in this case we can let anything be you and let anything be V. So from our right hand side of our equation. If we let you be equal to X2 and V be equal to e to the negative Y, then the derivative of u will be equal to 2 X, while the derivative of e to the negative Y is going to be equal to the derivative of the power, so that's gonna be -1 multiplied by DD X. So that we could write it as negative D the X multiplied by E to the negative Y, OK? So with that in mind, for our right hand side, when it, when it gets to the time when we'll need the poor rule, we can the product rule, sorry, we can use these to help us simplify. So now differentiating our function. Implicitly OK. Then we're going to have the derivative of y with respect to X on the left hand side and on the right hand side by applying the power rule differentiating X2 multiplied by e to the negative Y, we're going to have UV, so that's 2 X multiplied by e to the negative Y. Plus UV Prime. Negative DUIDX E to the negative Y multiplied by X2. Now we can go ahead and try to simplify the solve for DUIDX. Now, what this means, we're gonna have 2 XE to the negative Y minus X2 DUIDX E to the negative Y. And now if we. Bring that over to the left hand side. Then we're gonna have DUIDX plus X square. Let me just clean that up here. Plus X squared. OK, plus X 2 D Y D X E to the negative Y. OK. And now factoring on DUIDX. Divide the X multiplied by 1 + x2e the negative Y equals 2 xe to negative Y. This means then that DUIDX. It is going to be equal to 2 Xe negative Y divided by 1 + X2e the negative Y. Now, could we simplify this any further? Well, we could try to factor out E to the negative Y from the denominator and numerator, OK? Let me make some space over here. Now, to do that, OK. Then We can write in our denominator. Either the negative way. And no, we factor E to the negative Y from 1, it's gonna leave us with 1 divided by E to the negative Y plus X2. OK. For when we factor E to the negative Y from X2 E to the negative Y and no E to the negative Y will cancel. They give us divide the X equal to 2 X divided by 1. Divided by E to the negative Y plus X squared. Now we could take this a step further because what we know from the rules of powers is that for our reciprocality here we could rewrite this as E to the negative negative Y. OK, we could bring it up from the denominator. And no, that's gonna be either the negative negative Y or E to the Y plus X2, or we could even write it as 2 X divided by X2 + E to the power of Y, OK? So this is what DYD X will be equal to. And if we compare that to our answer choices, then we should find that option D is the correct answer. Thanks a lot for watching everyone. I hope this video helped.

27–40. Implicit differentiation Use implicit differentiation to find dy/dx.y = xe^y

Watch next

27–40. Implicit differentiation Use implicit differentiation to find dy/dx.
y = xe^y