Question

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

$y=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+...+\frac{x^n}{n!}=e^x$ (As it is expansion of $e^x$)

So, $y=e^x\Rightarrow \frac{dy}{dx}=e^x=y$