Derivative of Exponential Function

Theorem

Let $\exp$ be the exponential function.

Then:

$D_x \left({\exp x}\right) = \exp x$

Corollary 1

Let $c \in \R$.

Then:

$D_x \left({\exp \left({c x}\right)}\right) = c \exp \left({c x}\right)$

Corollary 2

Let $a \in \R: a > 0$.

Let $a^x$ be $a$ to the power of $x$.

Then:

$D_x \left({a^x}\right) = a^x \ln a$

Proof 1

We have:

	\(\displaystyle \)	\(\displaystyle D_x \left({\exp x}\right)\)	\(=\)	\(\displaystyle \lim_{h \to 0} \frac {\exp \left({x+h}\right) - \exp x} h\)	\(\displaystyle \)	by definition of derivative
	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \lim_{h \to 0} \frac {\exp x \left({\exp h - 1}\right)} h\)	\(\displaystyle \)	Exponent of Sum

From one of the definitions of the exponential function,

	\(\displaystyle \)	\(\displaystyle \frac{\exp h - 1} h\)	\(=\)	\(\displaystyle \frac{\lim_{n \to \infty} \left({1 + \frac h n}\right)^n - 1} h\)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \frac{\lim_{n \to \infty} \sum_{k=0}^n {n \choose k} 1^{n-k} \left({\frac h n}\right)^k - 1} h\)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \lim_{n \to \infty} \frac{\sum_{k=0}^n {n \choose k} 1^{n-k} \left({\frac h n}\right)^k - 1} h\)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \lim_{n \to \infty} {n \choose 1} \frac 1 n + \sum_{k=2}^n {n \choose k} \frac {h^{k-1} } {n^k}\)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle 1 + \lim_{n \to \infty} \sum_{k=2}^n {n \choose k} \frac{h^{k-1} } {n^k}\)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle 1 + h \lim_{n \to \infty} \sum_{k=2}^n {n \choose k} \frac{h^{k-2} } {n^k}\)	\(\displaystyle \)

The right summand converges to zero as $h$ gets arbitrarily small, and so:

$\displaystyle \lim_{h \to 0}\frac{\exp h - 1} h = 1$

From the Multiple Rule for Limits of Functions:

$\displaystyle \lim_{h \to 0} \frac {\exp x \left({\exp h - 1}\right)} {h} = \exp x \left({\lim_{x \to 0} \frac {\exp h - 1} h}\right)$

The result follows.

$\blacksquare$

Proof 2

We use the fact that the exponential function is the inverse of the natural logarithm function:

$y = e^x \iff x = \ln y$

	\(\displaystyle \)	\(\displaystyle \dfrac {\mathrm d x} {\mathrm d y}\)	\(=\)	\(\displaystyle \dfrac 1 y\)	\(\displaystyle \)	Derivative of Natural Logarithm Function
	\(\displaystyle \implies\)	\(\displaystyle \dfrac {\mathrm d y} {\mathrm d x}\)	\(=\)	\(\displaystyle \dfrac {1} {1 / y}\)	\(\displaystyle \)	Derivative of an Inverse Function
	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle y\)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle e^x\)	\(\displaystyle \)

$\blacksquare$

Proof 3

	\(\displaystyle \)	\(\displaystyle D_x (\ln e^x)\)	\(=\)	\(\displaystyle D_x (x)\)	\(\displaystyle \)	Exponential of Natural Logarithm
	\(\displaystyle \implies\)	\(\displaystyle \frac{1}{e^x}D_x (e^x)\)	\(=\)	\(\displaystyle 1\)	\(\displaystyle \)	Chain rule, Derivatives of Natural Log and Identity functions.
	\(\displaystyle \implies\)	\(\displaystyle D_x (e^x)\)	\(=\)	\(\displaystyle e^x\)	\(\displaystyle \)	multiply both sides by $e^x$

$\blacksquare$

Comment

That $D_x \exp x = \exp x$ can be used to define the exponential function.

See Equivalence of Exponential Definitions.

Sources

• Murray R. Spiegel: Mathematical Handbook of Formulas and Tables (1968): $13.29$
• K.G. Binmore: Mathematical Analysis: A Straightforward Approach (1977)... (previous)... (next): $\S 14.4$
• Roland E. Larson,  Robert P. Hostetler and Bruce H. Edwards: Calculus: 8th Edition (2005): $\S 5.4$
• For a video presentation of the contents of this page, visit the Khan Academy.