Question

Mark the correct alternative in each of the following:

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

(a) y + 1 (b) y − 1 (c) y (d) y²

Answer 1

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

Differentiating both sides with respect to x, we get

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

$\therefore \frac{dy}{dx}=y$

Hence, the correct answer is option (c).