Question

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

( \frac{dy}{dx} = )

Explanation:

Step1: Use the chain rule

Let \(u = \cot x-\csc x\), then \(y = u^{-1}\). The chain rule states that \(\frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dx}\). First, find \(\frac{dy}{du}\): \(\frac{dy}{du}=-u^{-2}=-\frac{1}{u^{2}}\).

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

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

Step3: Combine using the chain rule

Substitute \(u = \cot x-\csc x\) and \(\frac{du}{dx}\) into \(\frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dx}\). We get \(\frac{dy}{dx}=-\frac{1}{(\cot x - \csc x)^{2}}\cdot\csc x(\cot x-\csc x)=-\frac{\csc x}{\cot x - \csc x}\). Multiply numerator and denominator by \(- 1\): \(\frac{dy}{dx}=\frac{\csc x}{\csc x-\cot x}\).
Another way:
Simplify \(y = (\cot x-\csc x)^{-1}=\frac{1}{\cot x-\csc x}=\frac{1}{\frac{\cos x}{\sin x}-\frac{1}{\sin x}}=\frac{\sin x}{\cos x - 1}\)
Using the quotient rule \(\frac{d}{dx}(\frac{f(x)}{g(x)})=\frac{f^{\prime}(x)g(x)-f(x)g^{\prime}(x)}{g(x)^{2}}\), where \(f(x)=\sin x\), \(f^{\prime}(x)=\cos x\), \(g(x)=\cos x - 1\), \(g^{\prime}(x)=-\sin x\)
\(\frac{dy}{dx}=\frac{\cos x(\cos x - 1)-\sin x(-\sin x)}{(\cos x - 1)^{2}}=\frac{\cos^{2}x-\cos x+\sin^{2}x}{(\cos x - 1)^{2}}\)
Since \(\sin^{2}x+\cos^{2}x = 1\), then \(\frac{dy}{dx}=\frac{1-\cos x}{(\cos x - 1)^{2}}=\frac{1}{1 - \cos x}\)
Also, \(\frac{\csc x}{\csc x-\cot x}=\frac{\frac{1}{\sin x}}{\frac{1}{\sin x}-\frac{\cos x}{\sin x}}=\frac{1}{1-\cos x}\)

Answer:

\(\frac{1}{1 - \cos x}\)