# What is the second derivative of f(x)=cos(1/x) ?

May 28, 2018

$f ' ' \left(x\right) = \frac{- \cos \left(\frac{1}{x}\right) - 2 x \sin \left(\frac{1}{x}\right)}{x} ^ 4$

#### Explanation:

Here,

$f \left(x\right) = \cos \left(\frac{1}{x}\right)$

Diff.w.r.t. $x$ , using Chain Rule :

$\implies f ' \left(x\right) = - \sin \left(\frac{1}{x}\right) \frac{d}{\mathrm{dx}} \left(\frac{1}{x}\right)$

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

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

Again diff.w.r.t. $x$, using Quotient Rule :

$\implies f ' ' \left(x\right) = \frac{{x}^{2} \frac{d}{\mathrm{dx}} \left(\sin \left(\frac{1}{x}\right)\right) - \sin \left(\frac{1}{x}\right) \frac{d}{\mathrm{dx}} \left({x}^{2}\right)}{{x}^{2}} ^ 2$

$\implies f ' ' \left(x\right) = \frac{{x}^{2} \cdot \cos \left(\frac{1}{x}\right) \left(- \frac{1}{x} ^ 2\right) - \sin \left(\frac{1}{x}\right) \cdot 2 x}{x} ^ 4$

$\implies f ' ' \left(x\right) = \frac{- \cos \left(\frac{1}{x}\right) - 2 x \sin \left(\frac{1}{x}\right)}{x} ^ 4$