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

$\textcolor{red}{f ' \left(x\right) = \frac{4 x \cdot \sin {x}^{2} \cdot \cos {x}^{2} + 3 \cdot \sin x \cdot {\cos}^{2} x}{2 \sqrt{{\sin}^{2} {x}^{2} - {\cos}^{3} x}}}$

#### Explanation:

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

Using $\frac{d}{\mathrm{dx}} \left(\sqrt{u}\right) = \frac{1}{2 \sqrt{u}} \cdot \frac{d}{\mathrm{dx}} \left(u\right)$

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

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

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

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

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

$\textcolor{red}{f ' \left(x\right) = \frac{4 x \cdot \sin {x}^{2} \cdot \cos {x}^{2} + 3 \cdot \sin x \cdot {\cos}^{2} x}{2 \sqrt{{\sin}^{2} {x}^{2} - {\cos}^{3} x}}}$

God bless....I hope the explanation is useful.