Use implicit differentiation to find dy/dx.
sin xy = x+y...

Question

Use implicit differentiation to find dy/dx.

sin xy = x+y

Accepted Answer

Hi, everyone, let's take a look at this practice problem dealing with implicit differentiation. This problem says to bind the derivative of Y with respect to X by implicit differentiation, and we're given the equation, cosine of X multiplied by Y is equal to X2 plus Y2. So we need to find our derivative of Y with respect to X, and we're going to be using implicit differentiation. So we're going to take our equation here, which is cosine of X multiplied by Y is equal to X2 + Y2, and we're going to take the derivative with respect to X of both sides of this equation. And so, in doing so, on the left hand side, we will have the derivative with respect to X of cosine of X multiplied by Y. So when we take this derivative, we will have minus sine. Of X multiplied by Y. And this gets multiplied by the derivative with respect to X of X multiplied by Y. So here we'll have to use our product rule. We will have this being multiplied by the quantity, and here we'll recall our product rule. We'll take the derivative of our first term, which is going to be X, with respect to X, that's gonna give us 1, and that gets multiplied by our second term, which is Y. Then we'll have plus our first term, which is X, multiplied by the derivative of our second term, which is the derivative of Y with respect to X. And this is going to be equal to, and on the right hand side, I'll have the derivative of X2d with respect to X, which is going to give me 2 X. And then then we'll have plus the derivative of Y2d with respect to X, which is going to be equal to 2 Y multiplied by the derivative of Y with respect to X. So here we've just applied our chain rule. So, we need to solve this for the derivative of Y with respect to X. So the first step is to distribute our minus sign of X multiplied by Y. In doing so, I'll have minus Y multiplied by sine of X multiplied by Y. Then we'll have minus X multiplied by the derivative of Y with respect to X, multiplied by the sine of X multiplied by Y is equal to 2 X plus 2y multiplied by the derivative of Y with respect to X. So we need to collect all of our derivative of Y with respect to X terms on one side. So we'll move them all to the left hand side. To do that, we're going to need to subtract to Y multiplied by the derivative of Y with respect to X from both sides of the equation, and we're also going to need to add Um, Y multiplied by the sign of X multiplied by Y to both sides. So, in doing so, we will have minus 2 Y. Multiplied by the derivative of Y with respect to X. Minus X multiplied by the derivative of Y with respect to X, multiplied by the sine of X multiplied by Y is equal to 2 x plus Y multiplied by the sign of X multiplied by Y. On the left hand side, I can factor out the derivative of Y with respect to X out of each term, that will have the derivative. With with respect to X, multiplying the quantity of minus 2 Y minus X multiplied by sine of X multiplied by Y. In quantity is equal to 2 X plus Y multiplied by sine of X multiplied by Y. Now we just need to divide over the quantity that is multiplying the derivative of Y with respect to X. In doing so, I will get the derivative of Y. With respect to X, is equal to The quantity of 2 x plus y multiplied by the sign of X multiplied by Y in quantity divided by the quantity of minus 2y minus X multiplied by the sine of X multiplied by Y in quantity. So we can factor out a minus 1 out of our denominator, and in doing so, we'll get the derivative of Y with respect to X. is equal to minus the quantity of 2 X plus Y multiplied by the sine of X multiplied by Y in quantity, divided by the quantity of 2y plus X multiplied by the sign of X multiplied by Y in quantity. So this here is our final answer for this problem, since we have um found our derivative of Y with respect to X by using implicit differentiation. So, I hope that this explanation has been useful, and I'll see you in the next video.

Use implicit differentiation to find dy/dx.sin xy = x+y

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Use implicit differentiation to find dy/dx.
sin xy = x+y