Question

If \(x\,e^{xy} = y+sin^2x\), then the value of \(\frac{dy}{dx}\) at \(x = 0\) is equal to

• A. \(-1\)
• B. \(1\)
• C. \(-2\)
• D. \(2\)

Hint:

Find y at x = 0 first, then differentiate both sides.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
Use implicit differentiation, using the product rule on \(xe^{xy}\) and the chain rule on \(e^{xy}\) and \(\sin^2x\).

Step 2: Key Formula or Approach:
At \(x = 0\): \(0 \cdot e^{0} = y + 0\), so \(y(0) = 0\).

Step 3: Detailed Explanation:
Differentiate \(xe^{xy} = y + \sin^2x\) with respect to \(x\):
\[ e^{xy} + xe^{xy}\left(y + x\frac{dy}{dx}\right) = \frac{dy}{dx} + 2\sin x\cos x \]
Put \(x = 0\), \(y = 0\): \(1 + 0 = \frac{dy}{dx} + 0\).
So \(\frac{dy}{dx} = 1\).

Final Answer:
\(\frac{dy}{dx} = 1\) at \(x = 0\), option (B). \[ \boxed{1} \]