If y =1+x 1! +x^2 2! +x^3 3! + then d y d x = - Vidyadip

MCQ

If $y =1+\frac{x}{1!}+\frac{x^2}{2!}+\frac{x^3}{3!}+\ldots$ then $\frac{d y}{d x}=$

A
$y^2$
B
$y + 1$
✓
$y$
D
$y - 1$

✓

Answer

Correct option: C.

$y$

Differentiating both sides with respect to $x$,
we get $\frac{d y}{d x}=\frac{d}{d x}\left(1+\frac{x}{1!}+\frac{x^2}{2!}+\frac{x^2}{3!}+\ldots\right)$
$=\frac{d}{d x}(1)+\frac{d}{d w}\left(\frac{x}{11}\right)+\frac{d}{d w}\left(\frac{x^2}{2!}\right)+\frac{d}{d w}\left(\frac{x^3}{3!}\right)+\frac{d}{d x}\left(\frac{x^4}{4!}\right)+\ldots$
$=\frac{d}{d x}(1)+\frac{1}{1!} \frac{d}{d x}(x)+\frac{1}{2!} \frac{d}{d w}\left(x^2\right)+\frac{1}{3!} \frac{d}{d w}\left(x^3\right)+\frac{1}{4!} \frac{d}{d w}\left(x^4\right)+\ldots$
$=0+\frac{1}{1!} \times 1+\frac{1}{2!} \times 2 \alpha+\frac{1}{3!} \times 3 \alpha^2+\frac{1}{4!} \times 4 \alpha^3+\ldots\left(y=\alpha^2 \Rightarrow \frac{d y}{d \alpha}=n \alpha^{n-1}\right)$
$=1+\frac{x}{1!}+\frac{x^2}{2!}+\frac{x^3}{3!}+\ldots\left[\frac{x}{n!}=\frac{1}{(n-1)!}\right]$
$=y$
$\therefore \frac{d y}{d x}=y$

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