Difference between revisions of "Derivative of the exponential function"

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==References==
 
==References==
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* {{BookReference|Handbook of mathematical functions|1964|Milton Abramowitz|author2=Irene A. Stegun|prev=findme|next=findme}}: 4.2.5
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[[Category:Theorem]]
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[[Category:Proven]]
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[[Category:Justify]]

Latest revision as of 00:10, 23 December 2016

Theorem

The following formula holds: $$\dfrac{\mathrm{d}}{\mathrm{d}z} e^z = e^z,$$ where $e^z$ denotes the exponential function.

Proof

By definition, $$e^z = \displaystyle\sum_{k=0}^{\infty} \dfrac{z^k}{k!}.$$ Term-by-term differentiation of this sum shows $$\begin{array}{ll} \dfrac{\mathrm{d}}{\mathrm{d}z} e^z &= \displaystyle\sum_{k=0}^{\infty} \dfrac{\mathrm{d}}{\mathrm{d}z} \left[ \dfrac{z^k}{k!} \right] \\ &=\displaystyle\sum_{k=1}^{\infty} \dfrac{z^{k-1}}{(k-1)!} \\ &=\displaystyle\sum_{k=0}^{\infty} \dfrac{z^k}{k!} \\ &=e^z, \end{array}$$ as was to be shown. █

References