Question

The derivative of \(\tan^{-1} (x^2)\) w.r.t x is

(A) \(\frac x{1 + x^4}\)

(B) \(\frac {2x}{1 + x^4}\)

(C) \(-\frac {2x}{1 + x^4}\)

(D) \(\frac {1}{1 + x^4}\)

Answer 1

Correct option is (B) \(\frac{2x}{1 + x^4}\)

\(y = \tan^{-1} (x^2)\)

\(\tan y= x^2\)

Differentiate both sides with respect to x to get

\(\frac d{dy}(\tan y). \frac{dy}{dx} = \frac d{dx} (x^2)\)

\(\sec^2 y. \frac{dy}{dx} = 2x\)

\(\frac{dy}{dx} = \frac{2x}{\sec^2 y}\)

\(\frac{dy}{dx} = \frac{2x}{1 + \tan^2 y} \quad [\because \sec^2y = 1 + \tan^2 y]\)

Now, replace \(\tan y\) with \(x^2\) to get

\(\frac{dy}{dx} = \frac{2x}{1 + x^4}\)