Solveeit Logo

Question

Question: What is the derivative of \( y = \cos \left( {xy} \right) \) ?...

What is the derivative of y=cos(xy)y = \cos \left( {xy} \right) ?

Explanation

Solution

Hint : In the given problem, we are required to differentiate y=cos(xy)y = \cos \left( {xy} \right) with respect to x. Since, y=cos(xy)y = \cos \left( {xy} \right) is an implicit function, so we will have to differentiate the function y=cos(xy)y = \cos \left( {xy} \right) with the implicit method of differentiation. So, differentiation of y=cos(xy)y = \cos \left( {xy} \right) with respect to x will be done layer by layer using the chain rule of differentiation as in the given function, we cannot isolate the variables x and y.

Complete step by step solution:
Consider, y=cos(xy)y = \cos \left( {xy} \right) .
Differentiating both sides of the equation with respect to x, we get,
ddx(y)=ddx(cos(xy))\Rightarrow \dfrac{d}{{dx}}\left( y \right) = \dfrac{d}{{dx}}\left( {\cos \left( {xy} \right)} \right)
Using chain rule of differentiation, we will first differentiate cos(xy)\cos \left( {xy} \right) with respect to xyxy and then differentiate xyxy with respect to x, we get,
ddx(y)=dd(xy)(cos(xy))×ddx(xy)\Rightarrow \dfrac{d}{{dx}}\left( y \right) = \dfrac{d}{{d\left( {xy} \right)}}\left( {\cos \left( {xy} \right)} \right) \times \dfrac{d}{{dx}}\left( {xy} \right)
dydx=sin(xy)×ddx(xy)\Rightarrow \dfrac{{dy}}{{dx}} = - \sin \left( {xy} \right) \times \dfrac{d}{{dx}}\left( {xy} \right)
Now, using product rule of differentiation, we have, ddx[f(x)g(x)]=f(x)ddxg(x)+g(x)ddxf(x)\dfrac{d}{{dx}}\left[ {f\left( x \right)g\left( x \right)} \right] = f\left( x \right)\dfrac{d}{{dx}}g\left( x \right) + g\left( x \right)\dfrac{d}{{dx}}f\left( x \right) . So, we get,
dydx=sin(xy)×[xdydx+yddx(x)]\Rightarrow \dfrac{{dy}}{{dx}} = - \sin \left( {xy} \right) \times \left[ {x\dfrac{{dy}}{{dx}} + y\dfrac{d}{{dx}}\left( x \right)} \right]
Now, we know that the derivative of x with respect to x is 11 . So, we get,
dydx=sin(xy)×[xdydx+y]\Rightarrow \dfrac{{dy}}{{dx}} = - \sin \left( {xy} \right) \times \left[ {x\dfrac{{dy}}{{dx}} + y} \right]
Now, opening the brackets, we get,
dydx=xsin(xy)dydxysin(xy)\Rightarrow \dfrac{{dy}}{{dx}} = - x\sin \left( {xy} \right)\dfrac{{dy}}{{dx}} - y\sin \left( {xy} \right)
dydx+xsin(xy)dydx=ysin(xy)\Rightarrow \dfrac{{dy}}{{dx}} + x\sin \left( {xy} \right)\dfrac{{dy}}{{dx}} = - y\sin \left( {xy} \right)
dydx(1+xsin(xy))=ysin(xy)\Rightarrow \dfrac{{dy}}{{dx}}\left( {1 + x\sin \left( {xy} \right)} \right) = - y\sin \left( {xy} \right)
dydx=ysin(xy)1+xsin(xy)\Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{ - y\sin \left( {xy} \right)}}{{1 + x\sin \left( {xy} \right)}}
Therefore, we get the derivative of y=cos(xy)y = \cos \left( {xy} \right) as ysin(xy)1+xsin(xy)\dfrac{{ - y\sin \left( {xy} \right)}}{{1 + x\sin \left( {xy} \right)}}.

Note : Implicit functions are those functions that involve two variables and the two variables are not separable and cannot be isolated from each other. Hence, we have to follow a certain method to differentiate such functions and also apply the chain rule of differentiation. We must remember the simple derivatives of basic functions to solve such problems.