Question

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

• A. y+1
  
• B. y-1
  
• C. y
  
• D. $y^2$

Answer 1

The correct option is C y


$y=\frac{x}{1!}+\frac{x^2}{2!}+\frac{x^3}{3!}+...$
Differentiating both sides with respect to x, we get
$\frac{dy}{dx}=\frac{d}{dx}(1+\frac{x}{1!}+\frac{x^2}{2!}+\frac{x^3}{3!}+...)$
$=\frac{d}{dx}\left(1\right)+\frac{d}{dx}\left(\frac{x}{1!}\right)+\frac{d}{dx}\left(\frac{x^2}{2!}\right)+\frac{d}{dx}\left(\frac{x^3}{3!}\right)\frac{d}{dx}\left(\frac{x^4}{4!}\right)+....$
$=\frac{d}{dx}\left(1\right)+\frac{1}{1!}\frac{d}{dx}\left(x\right)+\frac{1}{2!}\frac{d}{dx}\left(x^2\right)+\frac{1}{3!}\frac{d}{dx}\left(x^3\right)+\frac{1}{4!}\frac{d}{dx}\left(x^4\right)+...$
$=0+\frac{1}{11}\times 1+\frac{1}{2!}\times 2x+\frac{1}{3!}\times 3x^2+\frac{1}{4!}\times 4x^3+...$
$(y=x^n\Rightarrow \frac{dy}{dx}=n{x}^{n-1})$
$=1+\frac{x}{1!}+\frac{x^2}{2!}+\frac{x^3}{3!}+...$ $[\frac{n}{n!}=\frac{1}{(n-1)!}]$
$=y$
$\therefore \frac{dy}{dx}=y$
Hence, the correct answer is option (c).