How do you differentiate f(x) = sqrt(sin^3(4-x)  using the chain rule?

May 13, 2018

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{- 3 \cdot {\sin}^{2} \left(4 - x\right) \cdot \cos \left(4 - x\right)}{2 \sqrt{{\sin}^{3} \left(4 - x\right)}}$

Explanation:

show below

y= sqrt(sin^3(4-x)

let suppose:

$u = 4 - x$

$\frac{\mathrm{du}}{\mathrm{dx}} = - 1$

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

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

$y = \sqrt{r}$

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

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

$\frac{\mathrm{dy}}{\mathrm{dx}} = - 1 \cdot 3 \cdot {\sin}^{2} u \cdot \cos u \cdot \frac{1}{2 \sqrt{r}}$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{- 3 \cdot {\sin}^{2} u \cdot \cos u}{2 \sqrt{r}}$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{- 3 \cdot {\sin}^{2} \left(4 - x\right) \cdot \cos \left(4 - x\right)}{2 \sqrt{{\sin}^{3} \left(4 - x\right)}}$