# How do you differentiate f(x)=sqrtsin(2-x^3)  using the chain rule?

##### 1 Answer
Mar 19, 2016

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

#### Explanation:

Let $y = \sqrt{\sin} \left(2 - {x}^{3}\right)$

Differentiating w.r.t. $x$

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

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

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

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

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

This will be the differentiated function.