Question

If \( x = e^{\tan^{-1} \left( \frac{y-x^2}{x^2} \right)} \), then \( \frac{dy}{dx} \) at \( x = 1 \) is

• A. 1
• B. 0
• C. 2
• D. 3

Hint:

Isolating $y$ before differentiating often makes the derivative calculation much cleaner.

Answer 1

The Correct Option is D

Step 1: Rewrite Equation
$\log x = \tan^{-1}(\frac{y}{x^2} - 1) \implies \frac{y}{x^2} - 1 = \tan(\log x)$.
$y = x^2 (1 + \tan(\log x))$.
Step 2: Differentiate
Using product rule: $dy/dx = 2x(1 + \tan(\log x)) + x^2 (\sec^2(\log x) \cdot \frac{1}{x})$.
Step 3: Evaluate at x=1
$\log(1) = 0$, $\tan(0) = 0$, $\sec(0) = 1$.
$dy/dx = 2(1)(1+0) + (1)^2 (1 \cdot 1) = 2 + 1 = 3$.
Step 4: Conclusion
The slope is 3.
Final Answer:(D)