Question

If $\log (x +y)-2 x y=0$ then $y'(0)=$

• A. $2 y^{2}+1$
• B. $2 y^{2}-1$
• C. $0$
• D. $2 y^{2}$

Answer 1

Answer: (B)

$\log (x +y)-2 x y=0$
Differentiating with respect to ' $x$ ' we get,
$\frac{1}{x+ y}\left(1+\frac{d y}{d x}\right)-2\left(x \frac{d y}{d x}+y\right)=0$
$\Rightarrow \frac{1}{x +y}+\left(\frac{1}{x+ y}-2 x\right) \frac{d y}{d x}-2 y=0$
$\Rightarrow \frac{d y}{d x}=\frac{2 y-\frac{1}{x +y}}{\frac{1}{x +y}-2 x}$
$\left.\Rightarrow \frac{d y}{d x}\right|_{x=0}=\frac{2 y-\frac{1}{y}}{\frac{1}{y}}=2 y^{2}-1$