Question

If $y=1+\frac{x}{1!}+\frac{x^2}{2!}+\frac{x^3}{3!}+\dots \mathrm{Find}\frac{dy}{dx}$

Answer 1

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

$\frac{dy}{dx}=1+\frac{2x}{2!}+\frac{{3x}^2}{3!}+\frac{{4x}^3}{4!}+......$

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

$=e^x$