Question

If log(x + y) = 2xy, then $\frac{dy}{dx}$ at x = 0 is

• A. 1
• B. -1
• C. 2
• D. -2

Answer 1

Answer: (A)

1

Answer 2

Given the equation $\log(x + y) = 2xy$.
We need to find $\frac{dy}{dx}$ at $x = 0$.

First, find the value of $y$ when $x=0$. Substitute $x=0$ into the given equation:
$\log(0 + y) = 2(0)y$
$\log(y) = 0$
Since $\log(y) = 0$ implies $y = e^0$, we have $y = 1$.
So, we need to find $\frac{dy}{dx}$ at the point $(x, y) = (0, 1)$.

Next, differentiate the given equation implicitly with respect to $x$.
$\frac{d}{dx}(\log(x + y)) = \frac{d}{dx}(2xy)$

Using the chain rule on the left side:
$\frac{1}{x+y} \cdot \frac{d}{dx}(x+y) = \frac{1}{x+y} (1 + \frac{dy}{dx})$

Using the product rule on the right side:
$\frac{d}{dx}(2xy) = 2 \cdot \frac{d}{dx}(xy) = 2 (1 \cdot y + x \cdot \frac{dy}{dx}) = 2y + 2x \frac{dy}{dx}$

Equating the derivatives of both sides:
$\frac{1}{x+y} (1 + \frac{dy}{dx}) = 2y + 2x \frac{dy}{dx}$

Now, we need to find the value of $\frac{dy}{dx}$ at the point $(0, 1)$. Substitute $x=0$ and $y=1$ into the differentiated equation:
$\frac{1}{0+1} (1 + \left.\frac{dy}{dx}\right|_{(0,1)}) = 2(1) + 2(0) \left.\frac{dy}{dx}\right|_{(0,1)}$
$\frac{1}{1} (1 + \left.\frac{dy}{dx}\right|_{(0,1)}) = 2 + 0$
$1 (1 + \left.\frac{dy}{dx}\right|_{(0,1)}) = 2$
$1 + \left.\frac{dy}{dx}\right|_{(0,1)} = 2$
$\left.\frac{dy}{dx}\right|_{(0,1)} = 2 - 1$
$\left.\frac{dy}{dx}\right|_{(0,1)} = 1$

Thus, the value of $\frac{dy}{dx}$ at $x=0$ is 1.