Question

find the derivative of the function ( y = (csc x - cot x)^{-1} ).

( \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 \((\csc x+\cot x)\):

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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:

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