Derivative of Exponential Function/Proof 1
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Theorem
Let $\exp$ be the exponential function.
Then:
- $\map {\dfrac \d {\d x} } {\exp x} = \exp x$
Proof
\(\ds \map {\frac \d {\d x} } {\exp x}\) | \(=\) | \(\ds \lim_{h \mathop \to 0} \frac {\map \exp {x + h} - \exp x} h\) | Definition of Derivative | |||||||||||
\(\ds \) | \(=\) | \(\ds \lim_{h \mathop \to 0} \frac {\exp x \cdot \exp h - \exp x} h\) | Exponential of Sum | |||||||||||
\(\ds \) | \(=\) | \(\ds \lim_{h \mathop \to 0} \frac {\exp x \paren {\exp h - 1} } h\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \exp x \paren {\lim_{h \mathop \to 0} \frac {\exp h - 1} h}\) | Multiple Rule for Limits of Real Functions, as $\exp x$ is constant | |||||||||||
\(\ds \) | \(=\) | \(\ds \exp x\) | Derivative of Exponential at Zero |
$\blacksquare$
Sources
- 1953: L. Harwood Clarke: A Note Book in Pure Mathematics ... (previous) ... (next): $\text {II}$. Calculus: Differentiation: Standard Differential Coefficients