Question

If $y=1+\frac{x}{1!}+\frac{x^2}{2!}+\frac{x^3}{3!}+.....,\text{then}\frac{dy}{dx}=........$

Answer 1

Given: $y=1+\frac{x}{1!}+\frac{x^2}{2!}+\frac{x^3}{3!}+.....$

To find: $\frac{dy}{dx}$

Differentiate with respect to $x$

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

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

Hence the value of
$\frac{dy}{dx}=1+\frac{x}{1!}+\frac{x^2}{2!}+\frac{x^3}{3!}+.........=y$