Question

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

Answer 1

Given that,

$y=1+\frac{x}{1!}+\frac{x^2}{2!}+\frac{x^3}{3!}+......,$ …… (1)

On differentiation this equation with respect to x and we get,

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

$\frac{dy}{dx}=0+1+\frac{2x}{2\times 1}+\frac{3x^2}{3\times 2!}+\frac{4x^3}{4\times 3!}+.......$

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

$\frac{dy}{dx}=y$ by equation (1)

Then,

$\frac{dy}{dx}-y=0$

Hence, this is the answer.