Question

The second order derivative of $\sec^{-1}\left(\frac{1}{2x^2 - 1}\right)$ with respect to $\cos^{-1}(2x^2 - 1)$, where $0<x<\frac{1}{\sqrt{2}}$ is

• A. 0
• B. 1
• C. $\frac{1}{2}$
• D. -1

Hint:

Before jumping into messy chain rule calculations for "derivative of $f(x)$ with respect to $g(x)$", always check if $f(x)$ can be simplified or rewritten directly in terms of $g(x)$ using algebraic or trigonometric identities. Recognizing $u=v$ reduces a 5-minute problem to a 5-second one.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
We are required to find the second derivative of one function with respect to another, i.e., \[ \frac{d^2 u}{dv^2} \] where \[ u = \sec^{-1}\left(\frac{1}{2x^2 - 1}\right), v = \cos^{-1}(2x^2 - 1) \]
Step 2: Key Identity:
Using the identity: \[ \sec^{-1}\left(\frac{1}{z}\right) = \cos^{-1}(z), \text{for } |z| \leq 1 \]
Step 3: Simplification:
Let $z = 2x^2 - 1$. Then, \[ u = \sec^{-1}\left(\frac{1}{2x^2 - 1}\right) = \cos^{-1}(2x^2 - 1) \] But \[ v = \cos^{-1}(2x^2 - 1) \] Hence, \[ u = v \]
Step 4: First Derivative:
\[ \frac{du}{dv} = \frac{d}{dv}(v) = 1 \]
Step 5: Second Derivative:
\[ \frac{d^2u}{dv^2} = \frac{d}{dv}(1) = 0 \]
Step 6: Final Answer:
\[ \boxed{0} \]