Question

The derivative of \(\dfrac{1}{\sin x}\) with respect to \(\cos x\) is:

• A. \(\sec x \tan^2 x\)
• B. \(\cot x \csc^2 x\)
• C. \(\tan x \sec^2 x\)
• D. \(\csc x \cot^2 x\)

Hint:

Use \(\dfrac{du}{dv}=\dfrac{du/dx}{dv/dx}\) with \(u=\csc x\), \(v=\cos x\), then simplify using cot x and cosec x.

Answer 1

The Correct Option is B

Step 1: Let \(u = \dfrac{1}{\sin x} = \csc x\) and \(v=\cos x\). We want \(\dfrac{du}{dv} = \dfrac{du/dx}{dv/dx}\).

Step 2: Differentiate \(u\) with respect to \(x\): \(\dfrac{du}{dx} = -\csc x \cot x\).

Step 3: Differentiate \(v\) with respect to \(x\): \(\dfrac{dv}{dx} = -\sin x\).

Step 4: So \(\dfrac{du}{dv} = \dfrac{-\csc x \cot x}{-\sin x} = \dfrac{\csc x \cot x}{\sin x}\). Since \(\dfrac{\csc x}{\sin x} = \csc^2 x\), this becomes \(\cot x \csc^2 x\).

Answer: option (B).