Question

If $y=\left(1+x\right)\left(1+x^2\right)...\left(1+x^{2n}\right)$, then $\frac{dy}{dx}$ at $x=0$ is

• A. $1$
• B. $-1$
• C. $0$
• D. None of these

Answer 1

The correct option is A

$1$

Explanation for the correct option.

Step 1. Find the value of $y$ at $x=0$.

It is given that $y=\left(1+x\right)\left(1+x^2\right)...\left(1+x^{2n}\right)$, now substitute $x=0$.

$\begin{array}{rcl}{\overline{)y}}_{x=0}& =& \left(1+0\right)\left(1+0^2\right)...\left(1+0^{2n}\right)\\ & =& 1\times 1\times 1...\times 1\\ & =& 1\end{array}$

Step 2. Differentiate $y$ with respect to $x$.

In the equation $y=\left(1+x\right)\left(1+x^2\right)...\left(1+x^{2n}\right)$ take log on both sides and simplify.

$\begin{array}{rcl}y& =& \left(1+x\right)\left(1+x^2\right)...\left(1+x^{2n}\right)\\ & \Rightarrow & \ln y=\ln \left(\left(1+x\right)\left(1+x^2\right)...\left(1+x^{2n}\right)\right)\left[\ln \left(ab\right)=\ln a+\ln b\right]\\ & \Rightarrow & \ln y=\ln \left(1+x\right)+\ln \left(1+x^2\right)+...+\ln \left(1+x^{2n}\right)\end{array}$

Now differentiate both sides with respect to $x$.

$\begin{array}{rcl}\frac{d}{dx}\left(\ln y\right)& =& \frac{d}{dx}\left(\ln \left(1+x\right)+\ln \left(1+x^2\right)+...+\ln \left(1+x^{2n}\right)\right)\\ & \Rightarrow & \frac{1}{y}\frac{dy}{dx}=\frac{1}{1+x}+\frac{2x}{1+x^2}+...+\frac{2n{x}^{2n-1}}{1+x^{2n}}\\ & \Rightarrow & \frac{dy}{dx}=y\left(\frac{1}{1+x}+\frac{2x}{1+x^2}+...+\frac{2n{x}^{2n-1}}{1+x^{2n}}\right)\end{array}$

Step 3. Find the value of $\frac{dy}{dx}$at $x=0$.

Now in the equation $\begin{array}{rcl}\frac{dy}{dx}& =& y\left(\frac{1}{1+x}+\frac{2x}{1+x^2}+...+\frac{2n{x}^{2n-1}}{1+x^{2n}}\right)\end{array}$ substitute $x=0$ and ${\overline{)y}}_{x=0}=1$ to find the value of $\frac{dy}{dx}$at $x=0$.

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

Hence, the correct option is A.