Question

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

Answer 1

We have,

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

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

On differentiating both sides w.r.t $x$, we get

$\frac{dy}{dx}=0+1+\frac{2x}{2!}+\frac{3x^2}{3!}+......+\frac{(n-1)x^{n-2}}{(n-1)!}+\frac{n{x}^{n-1}}{n!}$

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

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

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

$\frac{dy}{dx}=y-\frac{x^n}{n!}$

Hence, this is the answer.