Question

If $\ln(x + y) = 2xy,$ then $y^{'}(0)$ =

• A. 1
• B. – 1
• C. 2
• D. 0

Answer 1

Answer: (A)

1

Answer 2

$\ln(x + y) = 2xy$ ⇒ $\frac{(1 + dy/dx)}{(x + y)} = 2\left( x\frac{dy}{dx} + y \right)$

⇒ $\frac{dy}{dx} = \frac{1 - 2xy - 2y^{2}}{2x^{2} + 2xy - 1}$ ⇒ $y^{'}(0) = \frac{1 - 2}{- 1} = 1$, at $x = 0,$ $y = 1$.