Question

Let $y=1+\frac{x}{1!}+\frac{x^2}{2!}+........\infty$ then find $\frac{dy}{dx}$.

Answer 1

Given,

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

or, $y=e^x$ [Since $e^x=1+\frac{x}{1!}+\frac{x^2}{2!}+......\infty$]

Now,

$\frac{dy}{dx}$$=e^x=1+\frac{x}{1!}+\frac{x^2}{2!}+......\infty$