In this page, we will see how to calculate the derivative of the exponential function using two equivalent forms to express the incremental ratio: for \( h \to 0 \) and for \( x \to x_0 \). Formally, as:
\[ \lim_{h \to 0}\frac{a^{x + h} - a^x}{h} \quad , \quad \lim_{x \to x_0}\frac{a^x - a^{x_0}}{x - x_0} \]
Table of Contents
Limit of the incremental ratio as \( h \to 0 \)
We calculate the derivative of the exponential function \(f(x) = a^x\) as the limit of the incremental ratio:
\[ f'(x_0) = \lim_{h \to 0} \frac{a^{x_0 + h} - a^{x_0}}{h} \]
We rewrite the term \( a^{x_0 + h} = a^{x_0} \cdot a^h \), so:
\[ f'(x_0) = \lim_{h \to 0} \frac{a^{x_0} \cdot a^h - a^{x_0}}{h} \]
Now we can factor out \( a^{x_0} \) from the numerator:
\[ f'(x_0) = a^{x_0} \lim_{h \to 0} \frac{a^h - 1}{h} \]
The remaining limit is:
\[ \lim_{h \to 0} \frac{a^h - 1}{h} = \ln(a) \]
Therefore, the derivative of the exponential function is:
\[ f'(x_0) = a^{x_0} \ln(a) \]
Limit of the incremental ratio as \( x \to x_0 \)
We apply the definition of the derivative to the function \(f(x) = a^x\), obtaining:
\[ f'(x_0) = \lim_{x \to x_0} \frac{a^x - a^{x_0}}{x - x_0} \]
We rewrite \( a^x = a^{x_0} \cdot a^{x - x_0} \), so:
\[ f'(x_0) = a^{x_0} \lim_{x \to x_0} \frac{a^{x - x_0} - 1}{x - x_0} \]
We introduce an auxiliary variable \( u = x - x_0 \) (though it's not necessary), because for \( x \to x_0 \), \( x - x_0 \to 0 \). In this way, the limit becomes:
\[ L = \lim_{x \to x_0} \frac{a^{x - x_0} - 1}{x - x_0} = \lim_{u \to 0} \frac{a^u - 1}{u} = \ln(a) \]
The value of \( L \) is the natural logarithm of the base \( a \), i.e., \( \ln(a) \). Therefore, the derivative of the exponential function is:
\[ f'(x) = a^x \cdot \ln(a) \quad , \quad \forall x \in \mathbb{R} \]