Question

If $\log (x+y)-2xy=0$ then $y^{\prime }(0)=$

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

Answer 1

Answer: (B)

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