# How do you differentiate sqrt(sin^3(1/x)  using the chain rule?

May 7, 2018

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

#### Explanation:

Firstly
suppose:
$u = \frac{1}{x}$

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

$r = {\sin}^{3} \left(u\right)$

$\frac{\mathrm{dr}}{\mathrm{du}} = 3 {\sin}^{2} \left(u\right) \cdot \cos \left(u\right)$

$y = \sqrt{r}$

$\frac{\mathrm{dy}}{\mathrm{dr}} = \frac{1}{2 \sqrt{r}}$

now

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\mathrm{dy}}{\mathrm{dr}} \cdot \frac{\mathrm{dr}}{\mathrm{du}} \frac{\mathrm{du}}{\mathrm{dx}}$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{2 \sqrt{r}} \cdot 3 {\sin}^{2} \left(u\right) \cdot \cos \left(u\right) \left[- \frac{1}{x} ^ 2\right]$

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