Question

If $y=(1+x)(1+x^2)(1+x^4)......(1+x^{2n})$ then the value of $\left(\begin{array}{c}\frac{dy}{dx}\end{array}\right)$ at $x=0$ is

• A. $0$
• B. $-1$
• C. $1$
• D. $2$

Answer 1

The correct option is C $1$

We have,$y=(1+x)(1+x^2)(1+x^4).......(1+x^4)$

take natural logarithm both sides

$\ln y=\ln (1+x)+\ln (1+x^2)+\ln (1+x^4)+................+\ln (1+x^{2n})$

Now differentiate both sides w.r.t. $x$

$\Rightarrow \frac{1}{y}\frac{dy}{dx}=\frac{1}{1+x}+\frac{2x}{1+x^2}+\frac{4x^3}{1+x^4}+......+\frac{2n{x}^{2n-1}}{1+x^{2n}}$

$\Rightarrow \frac{dy}{dx}=y(\frac{1}{x}+\frac{2x}{1+x^2}+...........+\frac{2n{x}^{2n-1}}{1+x^{2n}})$

Now substitute $x=0$ to get the required value

$\Rightarrow \frac{dy}{dx}{|}_{x=0}=1(1+0+0+......+0)=1$

Note that at $x=0$, value of $y$ is also $1$