Question

The derivative of \(f (cot x)\) with respect to \(g (cosec x)\) at \(x=\frac{π}{4}\) (where \(f'(1)=2.g'(\sqrt2)=4\)) is:

• A. \(\sqrt{2}\)
• B. 1
• C. \(\frac{1}{\sqrt{2}}\)
• D. \(\frac{1}{2\sqrt{2}}\)

Answer 1

The Correct Option is C

To solve for the derivative of \(f(\cot x)\) with respect to \(g(\csc x)\) at \(x=\frac{\pi}{4}\), use the chain rule for derivatives. The chain rule states that if you have functions \(u(x)\) and \(v(x)\), then the derivative of \(u\) with respect to \(v\) is given by:

\[\frac{du}{dv}=\frac{du/dx}{dv/dx}\]

Given:

• \(u(x)=f(\cot x)\) with \(f'(1)=2\)
• \(v(x)=g(\csc x)\) with \(g'(\sqrt{2})=4\)

First, calculate the derivatives of \(u(x)\) and \(v(x)\) with respect to \(x\):

\[\frac{du}{dx}=f'(\cot x)\cdot (-\csc^2 x)\]

\[\frac{dv}{dx}=g'(\csc x)\cdot (-\csc x \cdot \cot x)\]

At \(x=\frac{\pi}{4}\):

• \(\cot\left(\frac{\pi}{4}\right)=1\) so \(f'(1)=2\)
• \(\csc\left(\frac{\pi}{4}\right)=\sqrt{2}\) so \(g'(\sqrt{2})=4\)
• \(\csc^2\left(\frac{\pi}{4}\right)=2\)
• \(\cot\left(\frac{\pi}{4}\right)=1\)

Substitute the values in:

\[\frac{du}{dx}=2 \cdot (-2)=-4\]

\[\frac{dv}{dx}=4 \cdot (-\sqrt{2}) \cdot 1 = -4\sqrt{2}\]

Therefore, the derivative of \(f(\cot x)\) with respect to \(g(\csc x)\) is:

\[\frac{du}{dv}=\frac{-4}{-4\sqrt{2}}=\frac{1}{\sqrt{2}}\]

The correct answer is \(\frac{1}{\sqrt{2}}\).