A) y
B) \[y-1\]
C) \[y+1\]
D) None of these
Correct Answer: A
Solution :
\[y=1+x+\frac{{{x}^{2}}}{2!}+\frac{{{x}^{3}}}{3!}+......\infty \]Þ\[y={{e}^{x}}\] Differentiating with respect to x, we get \[\frac{dy}{dx}={{e}^{x}}=y\].You need to login to perform this action.
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