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

Explanation:

Step1: Use the chain rule

The chain rule states that if (y = u^{-1}) where (u=\csc x-\cot x), then (\frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dx}). First, find (\frac{dy}{du}): If (y = u^{-1}), then (\frac{dy}{du}=-u^{-2}) by the power rule ((x^n)^\prime=nx^{n - 1}).

Step2: Find (\frac{du}{dx})

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

Step3: Combine using the chain rule

(\frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dx}=-u^{-2}\cdot\csc x(\csc x - \cot x)). Substitute (u = \csc x-\cot x) back in: (\frac{dy}{dx}=-\frac{\csc x(\csc x - \cot x)}{(\csc x - \cot x)^{2}}=-\frac{\csc x}{\csc x - \cot x}). We can also simplify further. Multiply the numerator and denominator by ((\csc x+\cot x)): [ \begin{align*} -\frac{\csc x}{\csc x - \cot x}\cdot\frac{\csc x+\cot x}{\csc x+\cot x}&=-\frac{\csc x(\csc x+\cot x)}{\csc^{2}x-\cot^{2}x}\ \end{align*} ] Since (\csc^{2}x-\cot^{2}x = 1), we have (\frac{dy}{dx}=-\csc x(\csc x+\cot x)=-\csc^{2}x-\csc x\cot x)

Another approach: We know that (y = (\csc x-\cot x)^{-1}=\frac{1}{\csc x-\cot x}). Multiply numerator and denominator by ((\csc x+\cot x)) (y=\frac{\csc x+\cot x}{\csc^{2}x-\cot^{2}x}=\csc x+\cot x) Now, find the derivative. ((\csc x)^\prime=-\csc x\cot x) and ((\cot x)^\prime=-\csc^{2}x) (\frac{dy}{dx}=-\csc x\cot x-\csc^{2}x)

Answer:

(-\csc^{2}x-\csc x\cot x)