Question

If $y=e^x.e^{x^2}.e^{x^3}.....e^{x^n}....,$ for $0<x<1,$ then $\frac{dy}{dx}$ at $x=\frac{1}{2}$ is

• A. e
• B. 4e
• C. 2e
• D. 3e
• E. 5e

Answer 1

Answer: (B)

Given, $y=e^x.e^{x^2}.e^{x^3}.....e^{x^n}.....$
$\Rightarrow$ $y=e^{(x+x^2+....+\infty )}$
$\Rightarrow$ $y=e^{\frac{x}{1-x}}$
$\Rightarrow$ $\frac{dy}{dx}=e^{\frac{x}{1-x}}\left[\frac{(1-x)1-x(-1)}{{(1-x)}^2}\right]$
$=e^{\frac{x}{1-x}}.\frac{1}{{(1-x)}^2}$ At $x=\frac{1}{2},$ ${\left(\frac{dy}{dx}\right)}_{x=\frac{1}{2}}=e^{\frac{\frac{1}{2}}{1-\frac{1}{2}}}.\frac{1}{{\left(1-\frac{1}{2}\right)}^2}$
$=4e$