Question

Let $y=\left(\frac{3^{x}-1}{3^{x}+1}\right) \sin x+\log _{e}(1+x), x>-1$. Then, at $x=0, \frac{d y}{d x}$ 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
$\frac{d y}{d x}=\frac{d}{d x}\left[\left(\frac{3^{x}-1}{3^{x}+1}\right) \sin x\right]+\frac{d}{d x} \log _{e}(1+x)$
$=\left(\frac{3^{x}-1}{3^{x}+1}\right) \frac{d}{d x} \sin x+\sin x \frac{d}{d x}\left(\frac{3^{x}-1}{3^{x}+1}\right)$
$+\frac{1}{1+x} \frac{d}{d x}(1+x)$
$=\left(\frac{3^{x}-1}{3^{x}+1}\right) \cos x+\sin x$
$\frac{\left(3^{x}+1\right) \frac{d}{d x}\left(3^{x}-1\right)-\left(3^{x}-1\right) \frac{d}{d x}\left(3^{x}+1\right)}{\left(3^{x}+1\right)^{2}}+\frac{1}{1+x}$
$=\left(\frac{3^{x}-1}{3^{x}+1}\right) \cos x+\sin x\left(3^{x}+1\right)\left(3^{x} \log _{e} 3\right)$
$\frac{-\left(3^{x}-1\right)\left(3^{x} \log _{e} 3\right)}{\left(3^{x}+1\right)^{2}}+\frac{1}{1+x}$
$\therefore \left(\frac{d y}{d x}\right)_{\text {at } x=0}=0+0+\frac{1}{1+0}=1$