Difference between revisions of "Derivative of the exponential function"

Revision as of 21:08, 6 June 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

• 1964: Milton Abramowitz and Irene A. Stegun: Handbook of mathematical functions ... (previous) ... (next): 4.2.5