The derivative f’ of the function f(x)=a^x is: f’(x) = ln(a) * a^x for any value of x and a > 0

Derivative of the Exponential Function a^x

The derivative $f’$ of the function $f(x)=a^x$ is:

\[\forall x \in ]-\infty, +\infty[ , f'(x) = \ln(a) \cdot a^{x}\]

where $a$ is a positive constant and $\ln(a)$ represents the natural logarithm of $a$.


Consider the function $f(x)=a^x$. Its derivative $f’(x)$ can be found using the definition of the derivative as a limit:

\[\begin{aligned} f^\prime(x) &=\lim _{h \rightarrow 0} \frac{a^{x+h}-a^{x}}{h} \\ &=\lim _{h \rightarrow 0} \frac{a^{x} \cdot a^{h}-a^{x}}{h} \\ &=a^{x}\cdot \lim _{h \rightarrow 0} \frac{a^{h}-1}{h} \end{aligned}\]

To find this limit, consider $a^h = e^{\ln(a^h)} = e^{h \cdot \ln(a)}$, therefore:

\[\begin{aligned} \lim _{h \rightarrow 0} \frac{a^{h}-1}{h} &= \lim _{h \rightarrow 0} \frac{e^{h \cdot \ln(a)}-1}{h} \\ &= \ln(a) \cdot \lim _{h \rightarrow 0} \frac{e^{h \cdot \ln(a)}-1}{h \cdot \ln(a)} \\ &= \ln(a) \cdot \lim _{h \rightarrow 0} \frac{e^{u}-1}{u} \quad \text{(with $u = h \cdot \ln(a)$)} \\ &= \ln(a) \cdot 1 \quad \text{(since the result of the limit is known to be 1)} \\ &= \ln(a) \end{aligned}\]

This finally gives us that:

\[f^\prime(x) = \ln(a) \cdot a^{x}\]

This result demonstrates that the derivative of the exponential function with base $a$ is equal to $\ln(a)$ multiplied by the function itself.