Question

If $y$ is a function of $x$ and $\log \left(x+y\right)=2xy$ then the value of $\frac{dy}{dx}$at $x=0$

• A. $1$
• B. $-1$
• C. $2$
• D. $0$

Answer 1

The correct option is A

$1$

Find the $\frac{dy}{dx}$:

Given that, $\log \left(x+y\right)=2xy....\left(1\right)$

Differentiate equation $\left(1\right)$with respect to $x.$

$\begin{array}{rcl}\frac{d\left[\log (x+y)\right]}{dx}& =& \frac{d\left(2xy\right)}{dx}\\ \frac{1}{x+y}\left[1+\frac{dy}{dx}\right]& =& 2x\frac{dy}{dx}+2y\\ \frac{1}{x+y}+\frac{1}{x+y}\frac{dy}{dx}& =& 2x\frac{dy}{dx}+2y\\ \frac{1}{x+y}\frac{dy}{dx}-2x\frac{dy}{dx}& =& 2y-\frac{1}{x+y}\\ \frac{dy}{dx}\left(\frac{1}{x+y}-2x\right)& =& 2y-\frac{1}{x+y}\\ \frac{dy}{dx}& =& \frac{2y-\frac{1}{x+y}}{\left(\frac{1}{x+y}-2x\right)}\end{array}$

For $x=0$,

$\log \left(y\right)=0$

$y=1$.

$\begin{array}{rcl}\frac{dy}{dx}& =& \frac{2-\frac{1}{0+1}}{\left(\frac{1}{0+1}-0\right)}\\ & =& 1\end{array}$

Hence, option A is correct.