Question

If $s = \sec^{-1} \left( \frac{1}{2x^2 - 1} \right)$ and $t = \sqrt{1 - x^2}$, then $\frac{ds}{dt}$ at $x = \frac{1}{2}$ is:

• A. $1$
• B. $2$
• C. $-2$
• D. $4$
• E. $-4$

Hint:

In parametric differentiation, if both $s$ and $t$ share a common denominator in their derivatives, they will cancel out. Focus on the numerators of $ds/dx$ and $dt/dx$ to find the final ratio quickly.

Answer 1

The Correct Option is D

Concept: This is a problem of differentiating one function with respect to another. We use the chain rule: $\frac{ds}{dt} = \frac{ds/dx}{dt/dx}$.

Step 1: Simplify the expression for $s$.
Recall that $\sec^{-1}(\frac{1}{z}) = \cos^{-1}(z)$. \[ s = \cos^{-1}(2x^2 - 1) \] Using the substitution $x = \cos \theta$, then $2\cos^2 \theta - 1 = \cos 2\theta$. \[ s = \cos^{-1}(\cos 2\theta) = 2\theta = 2\cos^{-1} x \] Differentiating $s$ with respect to $x$: \[ \frac{ds}{dx} = \frac{-2}{\sqrt{1 - x^2}} \]

Step 2: Differentiate $t$ with respect to $x$.
Given $t = \sqrt{1 - x^2}$: \[ \frac{dt}{dx} = \frac{1}{2\sqrt{1 - x^2}} \cdot (-2x) = \frac{-x}{\sqrt{1 - x^2}} \]

Step 3: Calculate $\frac{ds}{dt}$.
\[ \frac{ds}{dt} = \frac{ds/dx}{dt/dx} = \frac{-2 / \sqrt{1 - x^2}}{-x / \sqrt{1 - x^2}} = \frac{2}{x} \]

Step 4: Evaluate at $x = 1/2$.
\[ \left[ \frac{ds}{dt} \right]_{x=1/2} = \frac{2}{1/2} = 4 \]