Question

If y is a function of x and \( \log (x + y) = 2xy \), then the value of \( y'(0) = \).

• A. 2
• B. 0
• C. -1
• D. 1

Hint:

For implicit differentiation, solve for \( y' \) after finding a consistent point on the curve.

Answer 1

The Correct Option is D

Differentiate implicitly: \( \log (x + y) = 2xy \).
Left: \( \frac{1}{x + y} \cdot (1 + y') \). Right: \( 2(y + x y') \).
\[ \frac{1 + y'}{x + y} = 2y + 2x y'. \] At \( x = 0 \), find y: \( \log (0 + y) = 2 \cdot 0 \cdot y \Rightarrow \log y = 0 \Rightarrow y = 1 \).
Substitute \( x = 0 \), \( y = 1 \): \[ \frac{1 + y'}{0 + 1} = 2 \cdot 1 + 2 \cdot 0 \cdot y' \quad \Rightarrow \quad 1 + y' = 2 \quad \Rightarrow \quad y' = 1. \]