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

Question

Aicharya · Accepted Answer

Given $\begin{aligned} & y=1+x+\frac{x^{2}}{2 !}+\frac{x^{3}}{3 !}+\ldots \ 	herefore & \frac{d y}{d x}=e^{x}=y \end{aligned} \quad \Rightarrow y=e^{x}$

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

Solution