Question

If $y$ is a function of $x$ and $\log (x+y)=2xy$, then the value of $y^{\prime }(0)$ is

• A. 1
• B. - 1
• C. 2
• D. 0

Answer 1

Answer: (A)

Given that, $\log (x+y)=2xy....$(i)
$\therefore$ At $x=0,\Rightarrow \log (y)=0\Rightarrow y=1$
$\therefore$ To find $\frac{dy}{dx}$ at $(0,1)$
On differentiating Eq. (i) w.r.t. $x$, we get
$\frac{1}{x+y}(1+\frac{dy}{dx})=2x\frac{dy}{dx}+2y.1$
$\Rightarrow \frac{dy}{dx}=\frac{2y(x+y)-1}{1-2(x+y)x}$.
$\Rightarrow {\left(\frac{dy}{dx}\right)}_{(0,1)}=1$