Browse Questions

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

$\begin{array}{1 1} 1 \\ -1 \\ 2 \\ 0 \end{array}$

$\log(x+y)=2xy$
$\Rightarrow \large\frac{(1+dy/dx)}{(x+y)}=$$2.(x.\large\frac{dy}{dx}$$+y)$
$\Rightarrow \large\frac{dy}{dx}=\frac{1-2xy-2y^2}{2x^2+2xy-1}$
$\Rightarrow y'(0)=\large\frac{1-2}{-1}$
$\Rightarrow 1$ as at $x=0$,$y=1$
Hence (a) is the correct option