# What is the second derivative of f(x)=cos(x^2) ?

May 15, 2018

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

#### Explanation:

We use chain rule here

As $f \left(x\right) = \cos \left({x}^{2}\right)$

$\frac{\mathrm{df}}{\mathrm{dx}} = - \sin \left({x}^{2}\right) \times \frac{d}{\mathrm{dx}} {x}^{2} = - \sin \left({x}^{2}\right) \times 2 x = - 2 x \sin \left({x}^{2}\right)$

and $\frac{{d}^{2} f}{{\mathrm{dx}}^{2}} = - 1 \cdot 2 \sin \left({x}^{2}\right) - x \left[\cos \left({x}^{2}\right) \cdot 2 x\right]$

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

May 15, 2018

Here,

$f \left(x\right) = \cos {x}^{2}$

Diff.w.r.t., $x$, $\text{using "color(blue)"Chain Rule:}$

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

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

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

Again diff.w.r.t. $x$, $\text{ using "color(blue)"product Rule :}$,

$f ' ' \left(x\right) = - 2 \left[x \cos {x}^{2} \cdot 2 x + \sin {x}^{2}\right]$

$f ' ' \left(x\right) = - 2 \left[2 {x}^{2} \cos {x}^{2} + \sin {x}^{2}\right]$