# What is the derivative of the function y=sin(xy)?

## What is the derivative of the function $y = \sin \left(x y\right)$?

Oct 17, 2016

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{y \cos \left(x y\right)}{1 - x \cos \left(x y\right)}$

#### Explanation:

Using implicit differentiation, the product rule, and the chain rule, we get

$\frac{d}{\mathrm{dx}} y = \frac{d}{\mathrm{dx}} \sin \left(x y\right)$

$\implies \frac{\mathrm{dy}}{\mathrm{dx}} = \cos \left(x y\right) \left(\frac{d}{\mathrm{dx}} \left(x y\right)\right)$

$= \cos \left(x y\right) \left[x \left(\frac{d}{\mathrm{dx}} y\right) + y \left(\frac{d}{\mathrm{dx}} x\right)\right]$

$= \cos \left(x y\right) \left(x \frac{\mathrm{dy}}{\mathrm{dx}} + y\right)$

$= x \cos \left(x y\right) \frac{\mathrm{dy}}{\mathrm{dx}} + y \cos \left(x y\right)$

$\implies \frac{\mathrm{dy}}{\mathrm{dx}} - x \cos \left(x y\right) \frac{\mathrm{dy}}{\mathrm{dx}} = y \cos \left(x y\right)$

$\implies \frac{\mathrm{dy}}{\mathrm{dx}} \left(1 - x \cos \left(x y\right)\right) = y \cos \left(x y\right)$

$\therefore \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{y \cos \left(x y\right)}{1 - x \cos \left(x y\right)}$