# How do you differentiate f(x) = sin(x²)?

Jul 22, 2018

$\frac{\mathrm{df}}{\mathrm{dx}} = 2 x \cos \left({x}^{2}\right)$

#### Explanation:

Let $f \left(x\right) = \sin \left(g \left(x\right)\right)$ and $g \left(x\right) = {x}^{2}$

hence $\frac{\mathrm{df}}{\mathrm{dx}} = \frac{\mathrm{df}}{\mathrm{dg}} \times \frac{\mathrm{dg}}{\mathrm{dx}}$

= $\cos \left(g \left(x\right)\right) \times 2 x$

= $2 x \cos \left(g \left(x\right)\right)$

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

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

#### Explanation:

Given function:

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

Differentiating above function w.r.t. $x$ using chain rule:

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

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

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

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