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, $\frac{du}{dx}=\frac{d}{dx}(\csc x)+\frac{d}{dx}(\cot x)=-\csc x\cot x-\csc^{2}x=-\csc x(\cot x + \csc x)$.

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

Substitute $u = \csc x+\cot x$ and $\frac{du}{dx}=-\csc x(\csc x+\cot x)$ into $\frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dx}$. We have $\frac{dy}{dx}=-\frac{1}{(\csc x+\cot x)^{2}}\cdot(-\csc x(\csc x + \cot x))$.
Simplify the expression: $\frac{dy}{dx}=\frac{\csc x}{\csc x+\cot x}$.

Answer:

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