Question

If $y=1-x+\frac{x^2}{2!}-\frac{x^3}{3!}+\frac{x^4}{4!}...$, then $\frac{d^{2}y}{d{x}^2}$ is equal to

• A. $-x$
• B. $x$
• C. $y$
• D. $-y$

Answer 1

The correct option is C $y$

We have, $y=1-x+\frac{x^2}{2!}-\frac{x^3}{3!}+\frac{x^4}{4!}-....$

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

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