# How do you differentiate f(x)=sqrtcos(1/(2x)^3) using the chain rule?

Feb 14, 2018

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{3}{16} \sqrt{\sec \left(\frac{1}{8} {x}^{3}\right) \tan \left(\frac{1}{8} {x}^{3}\right)}$

#### Explanation:

Given:
f(x)=sqrt(cos(1/(2x)^3)

let $y = f \left(x\right)$

y=sqrt(cos(1/(2x)^3)

$y = \sqrt{s}$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{2 \sqrt{s}} \frac{\mathrm{ds}}{\mathrm{dx}}$

where,

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

$s = \cos t$

$\frac{\mathrm{ds}}{\mathrm{dx}} = - \sin t \frac{\mathrm{dt}}{\mathrm{dx}}$

where,

$t = \frac{1}{2 x} ^ 3$

$t = \frac{1}{u}$

$\frac{\mathrm{dt}}{\mathrm{dx}} = - \frac{1}{u} ^ 2 \frac{\mathrm{du}}{\mathrm{dx}}$

where,

$u = {\left(2 x\right)}^{3}$

$u = {v}^{3}$

$\frac{\mathrm{du}}{\mathrm{dx}} = 3 {v}^{2} \frac{\mathrm{dv}}{\mathrm{dx}}$

where,

$v = 2 x$

$\frac{\mathrm{dv}}{\mathrm{dx}} = 2$

$\frac{\mathrm{du}}{\mathrm{dx}} = 3 {v}^{2} \frac{\mathrm{dv}}{\mathrm{dx}}$

$\frac{\mathrm{du}}{\mathrm{dx}} = 3 {\left(2 x\right)}^{2} \left(2\right)$

$\frac{\mathrm{du}}{\mathrm{dx}} = 24 {x}^{2}$

$\frac{\mathrm{dt}}{\mathrm{dx}} = - \frac{1}{u} ^ 2 \frac{\mathrm{du}}{\mathrm{dx}}$

$\frac{\mathrm{dt}}{\mathrm{dx}} = - \frac{1}{{\left(2 x\right)}^{3}} ^ 2 \left(24 {x}^{2}\right)$

$\frac{\mathrm{dt}}{\mathrm{dx}} = - \frac{1}{64 {x}^{6}} \left(24 {x}^{2}\right)$

$\frac{\mathrm{dt}}{\mathrm{dx}} = - \frac{3}{8} {x}^{4}$

$\frac{\mathrm{ds}}{\mathrm{dx}} = - \sin t \frac{\mathrm{dt}}{\mathrm{dx}}$

$\frac{\mathrm{ds}}{\mathrm{dx}} = - \sin \left(\frac{1}{2 x} ^ 3\right) \left(- \frac{3}{8} {x}^{4}\right)$

$\frac{\mathrm{ds}}{\mathrm{dx}} = \frac{3}{8} \sin \left(\frac{1}{8} {x}^{3}\right)$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{2 \sqrt{\cos \left(\frac{1}{2 x} ^ 3\right)}} \left(\frac{3}{8} \sin \left(\frac{1}{8} {x}^{3}\right)\right)$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{3}{16} \frac{\sin \left(\frac{1}{8} {x}^{3}\right)}{\sqrt{\cos \left(\frac{1}{8} {x}^{3}\right)}}$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{3}{16} \sqrt{\frac{{\sin}^{2} \left(\frac{1}{8} {x}^{3}\right)}{\cos \left(\frac{1}{8} {x}^{3}\right)}}$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{3}{16} \sqrt{\sec \left(\frac{1}{8} {x}^{3}\right) \tan \left(\frac{1}{8} {x}^{3}\right)}$