Question

Let $y=\left(\frac{3^x-1}{3^x+1}\right)\sin x+{\log }_e(1+x),x>-1$. Then, at $x=0,\frac{dy}{dx}$ equals

• A. $1$
• B. $0$
• C. $-1$
• D. $-2$

Answer 1

The correct option is A $1$

Given,
$y=\left(\frac{3^x-1}{3^x+1}\right)\sin x+{\log }_e(1+x),x>-1$

Differentiating w.r.t. $x$, we get
$\frac{dy}{dx}=\frac{d}{dx}\left[\left(\frac{3^x-1}{3^x+1}\right)\sin x\right]+\frac{d}{dx}{\log }_e(1+x)$

$=\left(\frac{3^x-1}{3^x+1}\right)\frac{d}{dx}\sin x+\sin x\frac{d}{dx}\left(\frac{3^x-1}{3^x+1}\right)+\frac{1}{1+x}\frac{d}{dx}(1+x)$

$=\left(\frac{3^x-1}{3^x+1}\right)\cos x+\sin x$ $\frac{(3^x+1)\frac{d}{dx}(3^x-1)-(3^x-1)\frac{d}{dx}(3^x+1)}{(3^x+1{)}^2}+\frac{1}{1+x}$

$=\left(\frac{3^x-1}{3^x+1}\right)\cos x+\sin x(3^x+1)\left(3^x{\log }_{e}3\right)\frac{-(3^x-1)\left(3^x{\log }_{e}3\right)}{(3^x+1{)}^2}+\frac{1}{1+x}$

$\therefore {\left(\frac{dy}{dx}\right)}_{atx=0}=0+0+\frac{1}{1+0}=1$