Question

If \( y = \sec(\tan^{-1} x) \), then \( \frac{dy}{dx} \) is equal to

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

Hint:

Convert inverse trig expressions into algebraic form before differentiating.

Answer 1

Answer: (E)

Concept: Use chain rule and identity: \[ \sec(\tan^{-1}x) = \sqrt{1+x^2} \]

Step 1: Simplify function
Let \( \theta = \tan^{-1}x \Rightarrow \tan\theta = x \) \[ \sec\theta = \sqrt{1+\tan^2\theta} = \sqrt{1+x^2} \] Thus: \[ y = \sqrt{1+x^2} \]

Step 2: Differentiate
\[ \frac{dy}{dx} = \frac{1}{2}(1+x^2)^{-1/2}\cdot 2x \] \[ = \frac{x}{\sqrt{1+x^2}} \] But this corresponds to option (A), rechecking original: Actually derivative of \(\sec(\tan^{-1}x)\) via chain rule: \[ \frac{dy}{dx} = \sec(\tan^{-1}x)\tan(\tan^{-1}x)\cdot \frac{1}{1+x^2} \] \[ = \sqrt{1+x^2}\cdot x \cdot \frac{1}{1+x^2} \] \[ = \frac{x}{\sqrt{1+x^2}} \] Thus correct answer: \[ \boxed{\frac{x}{\sqrt{1+x^2}}} \]