Difference between revisions of "Derivative of the exponential function"

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(Created page with "<div class="toccolours mw-collapsible mw-collapsed"> <strong>Theorem:</strong> The following formula holds: $$\dfrac{\mathrm{d}}{\ma...")
 
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where $e^z$ denotes the [[exponential function]].
 
where $e^z$ denotes the [[exponential function]].
 
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<div class="mw-collapsible-content">
<strong>Proof:</strong> █  
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<strong>Proof:</strong> From the definition,
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$$e^z = \displaystyle\sum_{k=0}^{\infty} \dfrac{z^k}{k!}.$$
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[[Term-by-term differentiation]] of this sum shows
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$$\begin{array}{ll}
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\dfrac{\mathrm{d}}{\mathrm{d}z} e^z &= \displaystyle\sum_{k=0}^{\infty} \dfrac{\mathrm{d}}{\mathrm{d}z} \left[ \dfrac{z^k}{k!} \right] \\
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&=\displaystyle\sum_{k=1}^{\infty} \dfrac{z^{k-1}}{(k-1)!} \\
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&=\displaystyle\sum_{k=0}^{\infty} \dfrac{z^k}{k!} \\
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&=e^z,
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\end{array}$$
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as was to be shown. █  
 
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</div>
 
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Revision as of 05:56, 25 March 2016

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

Proof: From the 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. █