Question

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

• A. $e$
• B. $4e$
• C. $2e$
• D. $3e$

Answer 1

The correct option is B

$4e$

Solving the series:

Step 1: Simplify the expression

Given that, $y=e^x·e^{x^2}·e^{x^3}\dots e^{x^n}\dots .$

$e^x·e^{x^2}·e^{x^3}\dots e^{x^n}\dots .=e^{x+x^2+x^3+...}$ [using base same power add property]

Clearly, $x+x^2+x^3+...$ is an infinite geometric series with a common ratio $x.$

We know that sum of infinite geometric series is given by:

$S_{\infty }=\frac{a}{1-r};$

$$\begin{gathered} \Rightarrow x+x^2+x^3+...=\frac{x}{1-x} \\ \Rightarrow e^{x+x^2+x^3+...}=e^{\frac{x}{1-x}} \end{gathered}$$

Step 2: Find the $\frac{dy}{dx}$.

Now we have, $y=e^{\frac{x}{1-x}}$.

Differentiate both sides with respect to $x$ using chain rule.

$\begin{array}{rcl}\frac{dy}{dx}& =& {\mathrm{e}}^{\frac{\mathrm{x}}{1-\mathrm{x}}}\frac{\left[\right(1-\mathrm{x})+\mathrm{x}]}{(1-\mathrm{x}{)}^2}\\ & =& {\mathrm{e}}^{\frac{\mathrm{x}}{(1-\mathrm{x})}}\left[\frac{1}{(1-\mathrm{x}{)}^2}\right]\\ & =& \frac{{\mathrm{e}}^{{\displaystyle \frac{\mathrm{x}}{(1-\mathrm{x})}}}}{(1-\mathrm{x}{)}^2}\end{array}$

Substitute $x=\frac{1}{2}$we get:
$\begin{array}{rcl}\frac{\mathrm{dy}}{dx}& =& \frac{\mathrm{e}}{{\displaystyle \frac{1}{4}}}\\ & =& 4e\end{array}$

Hence, option (B) is correct.