If l n (x + y) = 2 x y ,then y^ ′ (0) is equal to

Given equation is $\ln( x + y ) = 2xy$...(1)

For $x = 0$

$\ln( 0 + y ) = 2.0.y = 0$

$⇒ \ln y = 0$

$⇒ y = 1$

Now, differentiating (1), we get,

$1\left(x + y\right)\left(1 + \dfrac{dy}{dx}\right) = 2y + 2x\dfrac{dy}{dx}$

At point (0,1), we get,

\left(\dfrac{1}{0 + 1}\right)\left(1 + \dfrac{dy}{dx}\right) =2 \left(1\right)$+ 2$\text{.} }}0\text{.}\left(\dfrac{dy}{dx}\right)$<br /><br />$⇒ 1 + \dfrac{dy}{dx} =2$<br /><br />$⇒ \dfrac{dy}{dx} =1= y^{'}( 0 )