Question

If \[ y=\sec^{-1}\left(\frac{1+x^2}{2x}\right) \quad \text{and} \quad x>1, \] then \[ \frac{dy}{dx}= \]

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

Hint:

Whenever you see expressions of the form \[ \frac{1+x^2}{2x}, \] try converting them into tangent half-angle identities. This frequently simplifies inverse trigonometric differentiation problems.

Answer 1

The Correct Option is B

Concept: The derivative of inverse trigonometric functions can often be simplified using algebraic identities before differentiation. A very important identity is: \[ \sec\theta+\tan\theta = x \] which can be manipulated to produce expressions involving: \[ \frac{1+x^2}{2x} \] Recognizing standard identities is the key idea in this problem.

Step 1: Rewrite the given expression in a recognizable form. We are given: \[ y=\sec^{-1}\left(\frac{1+x^2}{2x}\right) \] Observe that: \[ \frac{1+x^2}{2x} = \frac12\left(x+\frac1x\right) \] Now recall the identity: \[ \sec(\ln x) = \frac12\left(x+\frac1x\right) \] for \(x>0\). Hence, \[ \sec y=\frac12\left(x+\frac1x\right) \] which implies: \[ \cos y=\frac{2x}{1+x^2} \] But we know the standard trigonometric identity: \[ \cos(2\tan^{-1}t)=\frac{1-t^2}{1+t^2} \] and similarly, \[ \sec(2\tan^{-1}t)=\frac{1+t^2}{2t} \] Thus, \[ y=2\tan^{-1}x \]

Step 2: Differentiate the simplified expression. Differentiating: \[ y=2\tan^{-1}x \] we get: \[ \frac{dy}{dx} = 2\cdot\frac{1}{1+x^2} \] Therefore, \[ \boxed{ \frac{dy}{dx}=\frac{2}{1+x^2} } \]