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: Understanding the Question:
We need to find the derivative of an implicitly defined function.
Step 2: Key Formula or Approach:
Take the natural logarithm of both sides to simplify the expression before differentiating.
Step 3: Detailed Explanation:
\( x = e^{\tan^{-1} \left( \frac{y}{x^2} - 1 \right)} \implies \log_e x = \tan^{-1} \left( \frac{y}{x^2} - 1 \right) \).
First, find \( y \) at \( x = 1 \):
\( \log_e 1 = \tan^{-1}(y/1 - 1) \implies 0 = \tan^{-1}(y-1) \implies y - 1 = 0 \implies y = 1 \).
Differentiating with respect to \( x \):
\[ \frac{1}{x} = \frac{1}{1 + \left( \frac{y}{x^2} - 1 \right)^2} \times \frac{d}{dx} \left( \frac{y}{x^2} - 1 \right) \] \[ \frac{1}{x} = \frac{1}{1 + \left( \frac{y}{x^2} - 1 \right)^2} \times \left( \frac{x^2 y' - 2xy}{x^4} \right) \] At \( (1, 1) \):
\[ \frac{1}{1} = \frac{1}{1 + (1-1)^2} \times \left( \frac{1^2 y' - 2(1)(1)}{1^4} \right) \] \[ 1 = 1 \times (y' - 2) \implies y' = 3 \] Step 4: Final Answer:
The derivative \( dy/dx \) at \( x=1 \) is 3.