Question

Differential coefficient of see (\(tan^{-1} x\)) with respect to x is:

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

Answer 1

The Correct Option is C

To find the differential coefficient of \( \tan^{-1}x \) with respect to \( x \), we denote \( y = \tan^{-1}x \). The derivative of the inverse tangent function is given by:

\( \frac{dy}{dx} = \frac{1}{1 + x^2} \)

Now, to find the derivative of the secant of \( y \), namely \( \sec(y) = \sec(\tan^{-1}x) \), we first need to express \( \sec(y) \) in terms of \( x \):

Since \( y = \tan^{-1}x \), then \( \tan y = x \). We know the identity:

\( \sec^2 y = 1 + \tan^2 y \)

This implies:

\( \sec^2 y = 1 + x^2 \)

Taking the square root gives \( \sec y = \sqrt{1 + x^2} \).

Now differentiate \( \sec(y) \) with respect to \( x \):

\( \frac{d}{dx}[\sec(y)] = \sec(y)\tan(y)\cdot\frac{dy}{dx} \)

Substitute the values we've found:

\( = \sqrt{1 + x^2}\cdot x \cdot \frac{1}{1 + x^2} \)

Simplify:

\( = \frac{x}{\sqrt{1 + x^2}} \)

Therefore, the differential coefficient of \( \sec(\tan^{-1}x) \) with respect to \( x \) is:

\(\frac{x}{\sqrt{1 + x^2}}\)