Question

If y is a function of x and log (x + y) – 2xy = 0, then y′(0) is equal to –

• A. 1
• B. –1
• C. 2
• D. 0

Answer 1

Answer: (A)

1

Answer 2

We have : log (x + y) – 2xy = 0 … (1)

Diff. w.r.t. x, $\frac { 1 } { x + y }$ $\left( 1 + \frac{dy}{dx} \right)$ – 2x $\frac{dy}{dx}$ – 2y = 0

⇒ $\left( \frac { 1 } { x + y } - 2 x \right)$ $\frac{dy}{dx}$ = 2y – $\frac{1}{x + y}$

⇒ $\frac{dy}{dx}$ = $\frac{2y(x + y) - 1}{1 - 2x(x + y)}$.

When x = 0, from (1), log (y) = 0 ⇒ y = e⁰ = 1.

∴ $\left. \ \frac{dy}{dx} \right\rbrack_{x = 0}$ = $\frac{2(1)(0 + 1) - 1}{1 - 0}$ = $\frac{1}{1}$ = 1.