In this page we will see how to compute the derivative of the exponential function using two equivalent forms of the difference quotient: one with the variable \(h\), where \(h\to 0\), and one with the variable \(x\), where \(x\to x_0\).
Let \(a>0\), with \(a\neq 1\), and consider the exponential function:
\[ f(x)=a^x \]
The two forms of the difference quotient are:
\[ \lim_{h\to 0}\frac{a^{x+h}-a^x}{h} \qquad , \qquad \lim_{x\to x_0}\frac{a^x-a^{x_0}}{x-x_0} \]
Contents
Difference quotient for \( h\to 0 \)
We compute the derivative of the exponential function as the limit of the difference quotient:
\[ f'(x_0) = \lim_{h\to 0} \frac{a^{x_0+h}-a^{x_0}}{h} \]
Using the exponent rule:
\[ a^{x_0+h} = a^{x_0}\cdot a^h \]
we obtain:
\[ f'(x_0) = \lim_{h\to 0} \frac{a^{x_0}a^h-a^{x_0}}{h} \]
Factoring out \(a^{x_0}\):
\[ f'(x_0) = a^{x_0} \lim_{h\to 0} \frac{a^h-1}{h} \]
The remarkable limit of the exponential function is:
\[ \lim_{h\to 0} \frac{a^h-1}{h} = \ln(a) \]
Therefore:
\[ f'(x_0) = a^{x_0}\ln(a) \]
Difference quotient for \( x\to x_0 \)
We now apply the definition of derivative in the form:
\[ f'(x_0) = \lim_{x\to x_0} \frac{a^x-a^{x_0}}{x-x_0} \]
Rewrite \(a^x\) as:
\[ a^x = a^{x_0}\cdot a^{x-x_0} \]
Substituting:
\[ f'(x_0) = \lim_{x\to x_0} \frac{a^{x_0}a^{x-x_0}-a^{x_0}}{x-x_0} \]
Factoring out \(a^{x_0}\):
\[ f'(x_0) = a^{x_0} \lim_{x\to x_0} \frac{a^{x-x_0}-1}{x-x_0} \]
Introduce the auxiliary variable:
\[ u=x-x_0 \]
Since \(x\to x_0\), we have:
\[ u\to 0 \]
Hence the limit becomes:
\[ \lim_{u\to 0} \frac{a^u-1}{u} \]
By the remarkable limit of the exponential function:
\[ \lim_{u\to 0} \frac{a^u-1}{u} = \ln(a) \]
Therefore:
\[ f'(x_0) = a^{x_0}\ln(a) \]
In conclusion, the derivative of the exponential function is:
\[ f'(x) = a^x\ln(a) \qquad , \qquad \forall x\in\mathbb{R} \]