Derivative of the exponential function

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

• 1964: {{ #if: |{{{2}}}|Milton Abramowitz}}{{#if: Irene A. Stegun|{{#if: |, {{ #if: |{{{2}}}|Irene A. Stegun}}{{#if: |, [[Mathematician:{{{author3}}}|{{ #if: |{{{2}}}|{{{author3}}}}}]]{{#if: |, [[Mathematician:{{{author4}}}|{{ #if: |{{{2}}}|{{{author4}}}}}]]{{#if: |, [[Mathematician:{{{author5}}}|{{ #if: |{{{2}}}|{{{author5}}}}}]] and [[Mathematician:{{{author6}}}|{{ #if: |{{{2}}}|{{{author6}}}}}]]| and [[Mathematician:{{{author5}}}|{{ #if: |{{{2}}}|{{{author5}}}}}]]}}| and [[Mathematician:{{{author4}}}|{{ #if: |{{{2}}}|{{{author4}}}}}]]}}| and [[Mathematician:{{{author3}}}|{{ #if: |{{{2}}}|{{{author3}}}}}]]}}| and {{ #if: |{{{2}}}|Irene A. Stegun}}}}|}}: [[Book:Milton Abramowitz/Handbook of mathematical functions{{#if: |/Volume {{{volume}}}|}}{{#if: |/{{{edpage}}}}}|Handbook of mathematical functions{{#if: |: Volume {{{volume}}}|}}{{#if: |: {{{eddisplay}}}|{{#if: | ({{{ed}}} ed.)}}}}]]{{#if: | (translated by [[Mathematician:{{{translated}}}|{{ #if: |{{{2}}}|{{{translated}}}}}]])}}{{#if: |, {{{publisher}}}|}}{{#if: |, ISBN {{{isbn}}}|}}{{#if: findme | ... (previous)|}}{{#if: findme | ... (next)|}}{{#if: |: Entry: {{#if: |[[{{{entryref}}}|{{{entry}}}]]|{{{entry}}}}}|}}: 4.2.5