Question

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

• A. $y$
• B. $y+\frac{x^n}{n!}$
• C. $y-\frac{x^n}{n!}$
• D. $y-1-\frac{x^n}{n!}$

Answer 1

The correct option is C $y-\frac{x^n}{n!}$

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

Differentiate $y$ with respect to $x$, we get

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

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

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