# How do you differentiate f(x)=sqrtsin(2-4x)  using the chain rule?

Dec 25, 2015

Chain rule states that $\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\mathrm{dy}}{\mathrm{du}} \frac{\mathrm{du}}{\mathrm{dv}} \frac{\mathrm{dv}}{\mathrm{dx}}$

#### Explanation:

In this case, we can rename $y = \sqrt{u}$, $u = \sin v$ and $v = 2 - 4 x$, as we cannot derivate directly nor sqrt(sin(f(x)) neither $\sin \left(f \left(x\right)\right)$.

So, step-by-step solution:

$\frac{\mathrm{dy}}{\mathrm{du}} = \frac{1}{2 {u}^{\frac{1}{2}}}$

$\frac{\mathrm{du}}{\mathrm{dv}} = \cos v$

$\frac{\mathrm{dv}}{\mathrm{dx}} = - 4$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{2 {u}^{\frac{1}{2}}} \cos v \left(- 4\right)$

Substituting $u$:

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{2 \left(\sin v\right)} ^ \left(\frac{1}{2}\right) \cos v \left(- 4\right)$

Substituting $v$:

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{2 {\left(\sin \left(2 - 4 x\right)\right)}^{\frac{1}{2}}} \cos \left(2 - 4 x\right) \left(- 4\right)$

$\frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{2 \cos \left(2 - 4 x\right)}{\sqrt{\sin \left(2 - 4 x\right)}}$