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

Jun 23, 2016

$\frac{\mathrm{dy}}{\mathrm{dx}} = f ' \left(x\right)$=$- 6 x \sqrt{\sin} \left(2 - {x}^{2}\right) \cos \left(2 - {x}^{2}\right) .$

#### Explanation:

Let y=f(x)=sqrt(sin^3(2-x^2)

Put $u = {\sin}^{3} \left(2 - {x}^{2}\right) ,$ so, $y = \sqrt{u}$

Next, $u = {\sin}^{3} \left(2 - {x}^{2}\right) = {\left\{\sin \left(2 - {x}^{2}\right)\right\}}^{3} = {t}^{3} ,$ say, where $t = \sin \left(2 - {x}^{2}\right)$

Finally, take $2 - {x}^{2} = v$, so, $t = \sin v$

Thus, $y$ is a fun. of $u , u$ is a fun. of $t , t$ is a fun. of $v , v$ is a fun. of $x .$

Hence, by Chain Rule,

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\mathrm{dy}}{\mathrm{du}} \frac{\mathrm{du}}{\mathrm{dt}} \frac{\mathrm{dt}}{\mathrm{dv}} \frac{\mathrm{dv}}{\mathrm{dx}} \ldots \ldots \ldots . . \left(1\right)$

$y = \sqrt{u} \Rightarrow \frac{\mathrm{dy}}{\mathrm{du}} = \frac{1}{2 \sqrt{u}} .$

$u = {t}^{3} \Rightarrow \frac{\mathrm{du}}{\mathrm{dt}} = 3 {t}^{2.}$

$t = \sin v \Rightarrow \frac{\mathrm{dt}}{\mathrm{dv}} = \cos v .$

$v = 2 - {x}^{2} \Rightarrow \frac{\mathrm{dv}}{\mathrm{dx}} = - 4 x .$

Subing all these in $\left(1\right)$, we get,sqrt

$\frac{\mathrm{dy}}{\mathrm{dx}} = \left(\frac{1}{2 \sqrt{u}}\right) \left(3 {t}^{2}\right) \left(\cos v\right) \left(- 4 x\right)$
=-6xt^2cosv/sqrtu={-6xsin^2(2-x^2)cos(2-x^2)}/{sqrt(sin^3(2-x^2)}=$- 6 x \sqrt{\sin} \left(2 - {x}^{2}\right) \cos \left(2 - {x}^{2}\right) .$