Given equation is $\ln(x+y) = 2xy$. Find $\frac{dy}{dx}$ at point $(0,1)$.
Step-by-Step Solution
Key Concept: Use implicit differentiation on logarithmic equations and substitute the given point to find the derivative.
For $x=0$: $\ln(0+y) = 2(0)y = 0$, so $\ln y = 0$ which gives $y=1$. Differentiating $\ln(x+y) = 2xy$ implicitly: $\frac{1}{x+y}(1+\frac{dy}{dx}) = 2y + 2x\frac{dy}{dx}$. At point $(0,1)$: $\frac{1}{0+1}(1+\frac{dy}{dx}) = 2(1) + 2(0)\frac{dy}{dx}$. This simplifies to $1 + \frac{dy}{dx} = 2$, so $\frac{dy}{dx} = -1$ at $(0,1)$.
Correct Answer: -1