Question

If $y=\frac{1}{\sqrt[3]{3\csc x+\cot x}}$, find $\frac{dy}{dx}$

Answer 1

We have,

$y=\frac{1}{\sqrt[3]{3\csc x+\cot x}}$

$y=\frac{1}{(\csc x+\cot x{)}^{\frac{1}{3}}}$

$y=(\csc x+\cot x{)}^{-\frac{1}{3}}$

On differentiating w.r.t $x$, we get

$\frac{dy}{dx}=\frac{d}{dx}(\csc x+\cot x{)}^{-\frac{1}{3}}$

$\frac{dy}{dx}=-\frac{1}{3}(\csc x+\cot x{)}^{-\frac{4}{3}}\frac{d}{dx}(\csc x+\cot x)$

$\frac{dy}{dx}=-\frac{1}{3}(\csc x+\cot x{)}^{-\frac{4}{3}}(-\csc x\cot x-{\csc }^{2}x)$

$\frac{dy}{dx}=\frac{\csc x}{3}(\csc x+\cot x{)}^{-\frac{4}{3}}(\cot x+\csc x)$

$\frac{dy}{dx}=\frac{\csc x}{3}(\csc x+\cot x{)}^{-\frac{1}{3}}$

$\frac{dy}{dx}=\frac{\csc x}{3(\csc x+\cot x{)}^{\frac{1}{3}}}$

Hence, this is the answer.