Derivative of an Exponential Function
The derivative of an exponential function \( f(x)=a^x \) is \[ \frac{d}{dx}(a^x)=a^x\ln a \] where \( a>0 \) and \( a\neq 1 \).
One of the most remarkable properties of exponential functions is that their derivatives are also exponential functions. The only change is the appearance of the factor \(\ln a\).
This tells us that the rate of change of an exponential function is directly proportional to its value. As the function grows or decreases, its derivative follows the same exponential behavior.
- If the base is greater than 1, the function is increasing and its derivative is positive.
- If the base is between 0 and 1, the natural logarithm is negative, so the derivative is negative and the function is decreasing.
Note. When the base is Euler's number \( e \), we have \( \ln e =1 \), so the formula becomes \[ \frac{d}{dx}e^x=e^x \] This makes the natural exponential function unique: it is the only function whose derivative is exactly equal to the function itself.
Proof
Consider the exponential function
\[ f(x)=a^x \]
To find its derivative, start with the definition of the derivative as the limit of the difference quotient:
\[ f'(x)=\lim_{h \to 0}\frac{f(x+h)-f(x)}{h} \]
Substituting \( f(x)=a^x \) into the formula gives
\[ f'(x)=\lim_{h \to 0}\frac{a^{x+h}-a^x}{h} \]
Using the laws of exponents, we can rewrite \( a^{x+h} \) as \( a^x \cdot a^h \):
\[ f'(x)=\lim_{h \to 0}\frac{a^x(a^h-1)}{h} \]
Since \( a^x \) does not depend on \( h \), it can be factored out of the limit:
\[ f'(x)=a^x \cdot \lim_{h \to 0}\frac{a^h-1}{h} \]
The remaining limit is a well-known result known as the fundamental exponential limit:
\[ \lim_{h \to 0}\frac{a^h-1}{h}=\ln a \]
Substituting this result into the previous expression yields
\[ f'(x)=a^x \ln a \]
which is exactly the formula we wanted to prove.
An Example
Consider the function
\[ f(x)=2^x \]
Applying the derivative formula gives
\[ f'(x)=2^x \ln 2 \]
Since \( \ln 2 \approx 0.693 \), the derivative can also be written as
\[ f'(x)\approx 0.693 \cdot 2^x \]
Now evaluate the function at \( x=3 \):
\[ f(3)=2^3=8 \]
The corresponding value of the derivative is
\[ f'(3)=8 \cdot 0.693 \approx 5.54 \]
This means that when \( x=3 \), the function is increasing at a rate of approximately 5.54 units for every unit increase in \( x \).
The same approach can be used for any exponential function of the form \( a^x \), making this result one of the most important derivative formulas in calculus.
