Question

If $y=\left(\frac{3^x-1}{3^x+1}\right)\sin x+{\log }_e\left(1+x\right),x>-1$, then at $x=0$, $\frac{dy}{dx}$ is equal to

• A. $1$
• B. $0$
• C. $-1$
• D. $-2$

Answer 1

The correct option is A

$1$

Explanation for the correct option.

Find the value of $\frac{dy}{dx}$ at $x=0$.

Differentiate the equation $y=\left(\frac{3^x-1}{3^x+1}\right)\sin x+{\log }_e\left(1+x\right)$ with respect to $x$.

$\begin{array}{rcl}\frac{dy}{dx}& =& \frac{d}{dx}\left(\left(\frac{3^x-1}{3^x+1}\right)\sin x+{\log }_e\left(1+x\right)\right)\\ & =& \frac{3^x\log 3\left(3^x+1\right)-\left(3^x-1\right)3^x\log 3}{{\left(3^x+1\right)}^2}\sin x+\left(\frac{3^x-1}{3^x+1}\right)\cos x+\frac{1}{1+x}\end{array}$

Now to find the value of $\frac{dy}{dx}$ at $x=0$ substitute $0$ for $x$.

$\begin{array}{rcl}{\overline{)\frac{dy}{dx}}}_{x=0}& =& \frac{3^0\log 3\left(3^0+1\right)-\left(3^0-1\right)3^0\log 3}{{\left(3^0+1\right)}^2}\sin 0+\left(\frac{3^0-1}{3^0+1}\right)\cos 0+\frac{1}{1+0}\\ & =& \frac{1\times \log 3\times \left(1+1\right)-\left(1-1\right)\times 1\times \log 3}{{\left(1+1\right)}^2}\times 0+\left(\frac{1-1}{1+1}\right)\times 1+\frac{1}{1}\\ & =& 0+\frac{0}{1}+1\\ & =& 0+1\\ & =& 1\end{array}$

Thus at $x=0$, $\frac{dy}{dx}=1$.

Hence, the correct option is A.