Introduction
In calculus, an exponential function refers to any function that is a base with an exponent of some expression of $x$. Examples include $f(x) = 2^x$, $3^x$, $e^{x – 1}$, $5^{\sqrt{x + 1}}$, and so on. Derivatives of exponential functions often appear in calculus.
The Key Idea
The derivative of an exponential function $f(x) = a^x$ for some constant $a$ follows a simple formula. This formula is:
\[\begin{align*} \boxed{\dfrac{d}{dx}(a^x) = \ln a \cdot a^x}, \end{align*}\]
where $\ln a$ represents the constant $a$ inputted into the natural logarithm, or $\log_e a$. In other words, to find the derivative of $a^x$, we simply have to multiply the original function by $\ln a$. \\
\indent One important result of this formula arises when we notice that $\ln e = \log_e e = 1$. This means that
\[\begin{align*} \dfrac{d}{dx}(e^x) = \ln e \cdot e^x = e^x. \end{align*}\]
In other words, the derivative of $e^x$ is the function itself!
Generalization
All exponential functions can be represented in the form $a^{g(x)}$, where $a$ is some constant and $g(x)$ is some expression in terms of $x$. Essentially, all we’re saying is that exponential functions are some base, $a$, with an exponent that is some expression of $x$, which is $g(x)$.
To take the derivative, we will have to use the formula described above, along with the Chain Rule. The Chain Rule states that:
\[\begin{align*} \dfrac{d}{dx} (f(g(x))) = f'(g(x)) \cdot g'(x). \end{align*}\]
Now, set $f(x) = a^x$. Then, $f(g(x)) = a^{g(x)}$. In other words, taking the derivative of any exponential function is just an application of the Chain Rule — we’re taking the derivative of $f(g(x))$, where $f(x) = a^x$.
If we recall our previous formula, we will notice that $f'(g(x)) = \ln a \cdot a^{g(x)}$. Plugging this into our formula, we find:
\[\begin{align*} \dfrac{d}{dx}(a^{g(x)}) &= \dfrac{d}{dx}(f(g(x)) \\ &= f'(g(x)) \cdot g'(x) \\ &= \boxed{(\ln a \cdot a^{g(x)}) \cdot g'(x)}. \end{align*}\]
Example 1
Find the derivative of $h(x) = 2^{3x}$.
We want to apply our formula described above, so we want to match $2^{3x}$ with the general form $a^{g(x)}$. For this case, it becomes clear that $a = 2$ and $g(x) = 3x$. Plugging this into our formula, we have
\[\begin{align*} \dfrac{d}{dx}(2^{3x}) &= (\ln a \cdot a^{g(x)}) \cdot g'(x) \\ &= (\ln 2 \cdot 2^{3x}) \cdot (3x)’ \\ &= (\ln 2 \cdot 2^{3x}) \cdot 3 \\ &= 3\ln 2 \cdot 2^{3x}. \end{align*}\]
Example 2
Find the derivative of $e^{3x^2 + 1}$.
We want to match $e^{3x^2 + 1}$ with our general form $a^{g(x)}$. Now, it becomes clear that $a = e$ and $g(x) = 3x^2 + 1$. Plugging this into our formula, we have
\[\begin{align*} \dfrac{d}{dx}(e^{3x^2 + 1}) &= (\ln a \cdot a^{g(x)}) \cdot g'(x) \\ &= (\ln e \cdot e^{3x^2 + 1}) \cdot (3x^2 + 1)’. \end{align*}\]
Recall that $\ln e = \log_e e = 1$. Also, the derivative of $3x^2 + 1$ can be found using the Power Rule, giving us $(3x^2 + 1)’ = 6x$. Plugging this in, we find
\[\begin{align*} \dfrac{d}{dx}(e^{3x^2 + 1}) = (6x) \cdot e^{3x^2 + 1}. \end{align*}\]
Conclusion
It is helpful to break down the process into three steps.
- Match the exponential function with the form $a^{g(x)}$, helping us to identify $a$ and $g(x)$.
- Apply the formula, $\dfrac{d}{dx}(a^{g(x)}) = (\ln a \cdot a^{g(x)}) \cdot g'(x)$.
- Simplify the formula by evaluating $g'(x)$, and sometimes simplifying $\ln a$.
Following these three steps will ensure that you will arrive to the right answer.
