Question

find the derivative of the function y = (csc x - cot x)^(-1).
$\frac{dy}{dx}=square$

Explanation:

Step1: Apply chain - rule

Let $u = \csc x-\cot x$, so $y = u^{-1}$. By the chain - rule $\frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dx}$. First, find $\frac{dy}{du}$. Using the power rule, if $y = u^{-1}$, then $\frac{dy}{du}=-u^{-2}=-\frac{1}{u^{2}}$.

Step2: Find $\frac{du}{dx}$

We know that $\frac{d}{dx}(\csc x)=-\csc x\cot x$ and $\frac{d}{dx}(\cot x)=-\csc^{2}x$. So, $\frac{du}{dx}=\frac{d}{dx}(\csc x)-\frac{d}{dx}(\cot x)=-\csc x\cot x-(-\csc^{2}x)=\csc^{2}x - \csc x\cot x=\csc x(\csc x - \cot x)$.

Step3: Calculate $\frac{dy}{dx}$

Substitute $\frac{dy}{du}$ and $\frac{du}{dx}$ into the chain - rule formula: $\frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dx}=-\frac{1}{u^{2}}\cdot\csc x(\csc x - \cot x)$. Since $u = \csc x-\cot x$, we have $\frac{dy}{dx}=-\frac{\csc x(\csc x - \cot x)}{(\csc x - \cot x)^{2}}=-\frac{\csc x}{\csc x - \cot x}$.

Answer:

$-\frac{\csc x}{\csc x - \cot x}$