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

$y$

Step 1. Find the value of $\frac{dy}{dx}$.

Differentiate both sides of the equation $y=1-x+\frac{x^2}{2!}-\frac{x^3}{3!}+\frac{x^4}{4!}-...$ with respect to $x$.

$$\begin{gathered} \frac{dy}{dx}=\frac{d}{dx}\left(1-x+\frac{x^2}{2!}-\frac{x^3}{3!}+\frac{x^4}{4!}-...\right) \\ \Rightarrow \frac{dy}{dx}=0-1+\frac{2x}{2!}-\frac{3x^2}{3!}+\frac{4x^3}{4!}-... \\ \Rightarrow \frac{dy}{dx}=-1+x-\frac{x^2}{2!}+\frac{x^3}{3!}-... \\ \Rightarrow \frac{dy}{dx}=-\left(1-x+\frac{x^2}{2!}-\frac{x^3}{3!}+\frac{x^4}{4!}\right)\left[y=1-x+\frac{x^2}{2!}-\frac{x^3}{3!}+\frac{x^4}{4!}-..\right] \\ \Rightarrow \frac{dy}{dx}=-y...\left(1\right) \end{gathered}$$

Step 2. Find the value of $\frac{d^{2}y}{d{x}^2}$.

Now differentiate $\frac{dy}{dx}=-y$ with respect to $x$.

$\begin{array}{rcl}\frac{d^{2}y}{d{x}^2}& =& \frac{d}{dx}(-y)\\ & =& -\frac{dy}{dx}\\ & =& -(-y)\left[\mathrm{Using}\mathrm{equation}1\right]\\ & =& y\end{array}$

Hence, the correct option is C.