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

Answer: (A)

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
$\frac{dy}{dx}=\frac{d}{dx}\left[\left(\frac{3^x-1}{3^x+1}\right)\sin x\right]+\frac{d}{dx}{\log }_e(1+x)$
$=\left(\frac{3^x-1}{3^x+1}\right)\frac{d}{dx}\sin x+\sin x\frac{d}{dx}\left(\frac{3^x-1}{3^x+1}\right)$
$+\frac{1}{1+x}\frac{d}{dx}(1+x)$
$=\left(\frac{3^x-1}{3^x+1}\right)\cos x+\sin x$
$\frac{\left(3^x+1\right)\frac{d}{dx}\left(3^x-1\right)-\left(3^x-1\right)\frac{d}{dx}\left(3^x+1\right)}{{\left(3^x+1\right)}^2}+\frac{1}{1+x}$
$=\left(\frac{3^x-1}{3^x+1}\right)\cos x+\sin x\left(3^x+1\right)\left(3^x{\log }_{e}3\right)$
$\frac{-\left(3^x-1\right)\left(3^x{\log }_{e}3\right)}{{\left(3^x+1\right)}^2}+\frac{1}{1+x}$
$\therefore {\left(\frac{dy}{dx}\right)}_{\text{at }x=0}=0+0+\frac{1}{1+0}=1$