# How do you differentiate f(x)=x/cotsqrtx using the chain rule?

$\textcolor{red}{f ' \left(x\right) = \frac{2 \sqrt{x} \cdot \cot \sqrt{x} + x {\csc}^{2} \sqrt{x}}{2 \sqrt{x} \cdot {\cot}^{2} \sqrt{x}}}$

#### Explanation:

From the given $f \left(x\right) = \frac{x}{\cot \sqrt{x}}$

Use the Quotient Formula for finding derivatives

$\frac{d}{\mathrm{dx}} \left(\frac{u}{v}\right) = \frac{v \cdot \frac{d}{\mathrm{dx}} \left(u\right) - u \cdot \frac{d}{\mathrm{dx}} \left(v\right)}{v} ^ 2$

Let $u = x$ and $v = \cot \sqrt{x}$

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

$f ' \left(x\right) = \frac{\cot \sqrt{x} + x {\csc}^{2} \sqrt{x} \cdot \frac{1}{2 \sqrt{x}}}{\cot \sqrt{x}} ^ 2$

$\textcolor{red}{f ' \left(x\right) = \frac{2 \sqrt{x} \cdot \cot \sqrt{x} + x {\csc}^{2} \sqrt{x}}{2 \sqrt{x} \cdot {\cot}^{2} \sqrt{x}}}$

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