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Differentiation of Derivatives — Applied Maths STD 11 Science — Question

CBSE BoardEnglish MediumSTD 11 ScienceApplied MathsDifferentiation of Derivatives1 Mark
Question
If $y=1+\frac{x}{1!}+\frac{x^2}{2!}+\frac{x^3}{3!} \ldots$, then $\frac{d y}{d x}=$_________________
✓

Answer

$y$, because
Given that,
$\begin{aligned} y & =1+\frac{x}{1!}+\frac{x^2}{2!}+\frac{x^3}{3!} ...., \\ \frac{d y}{d x} & =0+\frac{1}{1!}+\frac{2 x}{2!}+\frac{3 x^2}{3!}+.... \\ & =1+\frac{x}{1!}+\frac{x^2}{2!}+\frac{x^3}{3!} .... =y .\end{aligned}$

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