Question

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

Answer 1

Hint: In the given problem, we are required to differentiate $ y = \cos \left( {xy} \right) $ with respect to x. Since, $ y = \cos \left( {xy} \right) $ is an implicit function, so we will have to differentiate the function $ y = \cos \left( {xy} \right) $ with the implicit method of differentiation. So, differentiation of $ 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 \left( {xy} \right) $ .
Differentiating both sides of the equation with respect to x, we get,
 $ \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 \left( {xy} \right) $ with respect to $ xy $ and then differentiate $ xy $ with respect to x, we get,
 $ \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) $
 $ \Rightarrow \dfrac{{dy}}{{dx}} = - \sin \left( {xy} \right) \times \dfrac{d}{{dx}}\left( {xy} \right) $
Now, using product rule of differentiation, we have, $ \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,
 $ \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 $ 1 $ . So, we get,
 $ \Rightarrow \dfrac{{dy}}{{dx}} = - \sin \left( {xy} \right) \times \left[ {x\dfrac{{dy}}{{dx}} + y} \right] $
Now, opening the brackets, we get,
 $ \Rightarrow \dfrac{{dy}}{{dx}} = - x\sin \left( {xy} \right)\dfrac{{dy}}{{dx}} - y\sin \left( {xy} \right) $
 $ \Rightarrow \dfrac{{dy}}{{dx}} + x\sin \left( {xy} \right)\dfrac{{dy}}{{dx}} = - y\sin \left( {xy} \right) $
 $ \Rightarrow \dfrac{{dy}}{{dx}}\left( {1 + x\sin \left( {xy} \right)} \right) = - y\sin \left( {xy} \right) $
 $ \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{ - y\sin \left( {xy} \right)}}{{1 + x\sin \left( {xy} \right)}} $
Therefore, we get the derivative of $ y = \cos \left( {xy} \right) $ as $ \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.