Question

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

• A. $0$
• B. $-1$
• C. $1$
• D. $2$

Answer 1

The correct option is C

$1$

Explanation for the correct answer:

$y=(1+x)(1+x^2)(1+x^4)...(1+x^{2n})$ …[given]

Taking natural logarithm on both sides we get

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

$\Rightarrow$$\ln y=\ln (1+x)+\ln (1+x^2)+\ln (1+x^4)...+\ln (1+x^{2n})$ $\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\left[\because \ln \left(a\times b\times c\right)=\ln a+\ln b+\ln c\right]$

Differentiating the above equation with respect to $x$ we get

$\Rightarrow$$\frac{1}{y}\frac{dy}{dx}=\frac{1}{1+x}+\frac{1}{1+x^2}\times 2x+\frac{1}{1+x^4}\times 4x^3+.....+\frac{1}{1+x^{2n}}\times 2n{x}^{2n-1}$ $\mathbf{.}\mathbf{.}\mathbf{.}\left[\because \frac{d}{\mathrm{dx}}\left(\ln x\right)=\frac{1}{x}\right]$

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

Substituting the value of $y$ we get

$\Rightarrow$ $\frac{dy}{dx}=(1+x)(1+x^2)(1+x^4)...(1+x^{2n})\left[\frac{1}{1+x}+\frac{2x}{1+x^2}+\frac{4x^3}{1+x^4}+.....+\frac{2n{x}^{2n-1}}{1+x^{2n}}\right]$

To find value of ${\left(\frac{dy}{dx}\right)}_{x=0}$ we substitute $x=0$ in

$\Rightarrow$${\left(\frac{dy}{dx}\right)}_{x=0}=\left(1\right)\left(1\right)\left(1\right)...\left(1\right)\left[\frac{1}{1}+0+0+.....+0\right]$

$\Rightarrow$${\left(\frac{dy}{dx}\right)}_{x=0}=1$

Hence, the value of ${\left(\frac{dy}{dx}\right)}_{x=0}$ is $1$,

Hence, option (C) is the correct answer.