# How do you find the derivative of f(x)=sin (1/x^2) using the chain rule?

Apr 1, 2018

${f}^{'} \left(x\right) = - \frac{2}{x} ^ 3 \cos \left(\frac{1}{x} ^ 2\right)$

#### Explanation:

Here,

$f \left(x\right) = \sin \left(\frac{1}{x} ^ 2\right) = \sin \left({x}^{-} 2\right)$

Using Chain Rule,we get

${f}^{'} \left(x\right) = \cos \left({x}^{-} 2\right) \frac{d}{\mathrm{dx}} \left({x}^{-} 2\right)$

$= \cos \left(\frac{1}{x} ^ 2\right) \left(- 2 {x}^{- 2 - 1}\right)$

$= - 2 \cos \left(\frac{1}{x} ^ 2\right) \left({x}^{-} 3\right)$

$= - \frac{2}{x} ^ 3 \cos \left(\frac{1}{x} ^ 2\right)$