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If $y=1+x+\frac{x^{2}}{2 !}+\frac{x^{3}}{3 !}+\ldots \ldots$ then $\frac{d y}{d x}=$
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Verified Answer
The correct answer is:
$y$
Given
$$
\begin{aligned}
& y=1+x+\frac{x^{2}}{2 !}+\frac{x^{3}}{3 !}+\ldots \\
\therefore & \frac{d y}{d x}=e^{x}=y
\end{aligned} \quad \Rightarrow y=e^{x}
$$
$$
\begin{aligned}
& y=1+x+\frac{x^{2}}{2 !}+\frac{x^{3}}{3 !}+\ldots \\
\therefore & \frac{d y}{d x}=e^{x}=y
\end{aligned} \quad \Rightarrow y=e^{x}
$$
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