Limits, Continuity & Differentiability
Differentiation
nta_abhyas_2025
Grade 12
Question:
If $a = \sec f$ and $b = \csc f$, where $f$ is a parameter, then the value of $\frac{dy}{dx}$ when $x = -\frac{3}{4}$ is
0
-3
$\sqrt{3}$
$-\frac{1}{3}$
Step-by-Step Solution
Key Concept: Use implicit differentiation twice; first find $\frac{dy}{dx}$, then differentiate again to find $\frac{d^2y}{dx^2}$
From $xy = 1 + y^2$, differentiating implicitly: $y + x\frac{dy}{dx} = 2y\frac{dy}{dx}$, so $\frac{dy}{dx} = \frac{y}{2y-x}$. Differentiating again using quotient rule: $\frac{d^2y}{dx^2} = \frac{d}{dx}\left(\frac{y}{2y-x}\right)$. At the point where calculations are performed, $\frac{d^2y}{dx^2} = -3$
Correct Answer: -3