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: Write $u=\dfrac{1}{\sin x}$ and $v=\cos x$, and differentiate $u$ directly using the quotient rule instead of the standard cosec derivative formula: $\dfrac{du}{dx} = \dfrac{0\cdot\sin x - 1\cdot\cos x}{\sin^2 x} = \dfrac{-\cos x}{\sin^2 x}$.

Step 2: Also $\dfrac{dv}{dx} = -\sin x$.

Step 3: Then $\dfrac{du}{dv} = \dfrac{du/dx}{dv/dx} = \dfrac{-\cos x/\sin^2 x}{-\sin x} = \dfrac{\cos x}{\sin^3 x}$.

Step 4: Rewrite this in terms of standard trig ratios: $\dfrac{\cos x}{\sin^3 x} = \left(\dfrac{\cos x}{\sin x}\right)\left(\dfrac{1}{\sin^2 x}\right) = \cot x \csc^2 x$.

Step 5: This independent quotient-rule computation matches option (B) exactly, confirming the result obtained via the direct cosec derivative formula.

\[\boxed{\cot x \csc^2 x}\]