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# Exponential derivative – Derivation, Explanation, and Example

In differential calculus, we’ll need to also establish a rule for**exponential derivative**. Our discussion will revolve around the formula for $\dfrac{d}{dx} a^x$ and $\dfrac{d}{dx} e^x$. Exponential functions have a wide range of applications in different STEM fields, so it’s essential to understand how its derivative behaves.

**$\boldsymbol{e^x}$**

*The derivative of an exponential function will be the function itself and a constant factor. A special case occurs for***$\boldsymbol{e^x}$**

*since the derivative is***In this article, we’ll understand how we could come up with the exponential functions’ derivative rules. We’ll also see how we can apply them to differentiate a wide range of functions. This is why it’s important to take a refresher on the following topics:**

*as well.*- Review how we derive derivative rules from the formal definition of derivatives.
- Understand the components of an exponential function.
- Apply the fundamental derivative rules with these new derivative rules.

**What is the derivative of an exponential function? **

Recall that an exponential function has a general form of $y = a^x$, where $a > 0$ but $a \neq 1$. We call $a$ the base, and $x$ is in the exponent part of the expression. For functions like this, **the derivative will be the function itself times the base’s natural logarithm**. \begin{aligned}\dfrac{d}{dx} a^x &= a^x \ln a\end{aligned} Let’s go ahead and observe how an exponential function and its derivative would look like when graphed on an $xy$-plane.

Derivative Rules
Base of $\boldsymbol{a}$ |
Derivative Rules
Base of $\boldsymbol{e}$ |

\begin{aligned}\dfrac{d}{dx} a^x &= a^x \ln a\end{aligned} | \begin{aligned}\dfrac{d}{dx} e^x &= e^x\end{aligned} |

\begin{aligned}\dfrac{d}{dx} a^{[g(x)]} &= a^{[g(x)]} \ln [g(x)] \cdot g’(x)\end{aligned} | \begin{aligned}\dfrac{d}{dx} e^{[g(x)]} &= e^{[g(x)]} \cdot g’(x)\end{aligned} |

**Proof of the derivative rule for exponential functions**

Recall that $\dfrac{d}{dx} f(x) = \lim_{h\rightarrow 0}\dfrac{f(x + h) – f(x)}{h}$, so we can use this to confirm the derivative that we’ve just learned for $y = a^x$.
- Use the product rule for exponents,$a^{m} \cdot a^n = a^{m+n}$, to factor $a^x$ from the numerator.
- Since $a^x$ is considered a constant for this expression, we can factor it out of $\lim_{h \rightarrow 0}$.
- Evaluate the limit by setting $h$ to $0$.

**How to find the derivative of an exponential function?**

The most important step in differentiating exponential functions is to **make sure that we’re actually working with exponential functions**. Make sure that the

**function has a constant base and**$\boldsymbol{x}$

**is found at the exponent.**Once we’ve confirmed that the function (or the composite function’s outer layer) has a form of either $y= a^x$ or $y = e^x$, we can then apply the derivative rule we’ve just learned.

- To find the function’s derivative, copy the original function.
- Multiply this with the natural logarithm of the base. (Skip this when working with $y= e^x$.)
- If working with composite functions, multiply the result with the derivative of the inner function.

\begin{aligned}\boldsymbol{\dfrac{d}{dx} a^x = a^x \ln a}\end{aligned} | \begin{aligned}\boldsymbol{\dfrac{d}{dx} e^x = e^x}\end{aligned} |

\begin{aligned}\dfrac{d}{dx} (5^x) &= 5^x \cdot \ln 5,\phantom{x}\color{DarkOrange}\text{Derivative of }a^x \end{aligned} | \begin{aligned}\dfrac{d}{dx} (6e^x) &= 6\cdot \dfrac{d}{dx}e^x,\phantom{x}\color{DarkOrange}\text{Constant Multiple Rule}\\&= 6\cdot{\color{DarkOrange}e^x},\phantom{x}\color{DarkOrange}\text{Derivative of }e^x \\&= 6e^x\end{aligned} |

\begin{aligned}\dfrac{d}{dx} (2^{3x}) &= {\color{DarkOrange}2^{3x} \cdot \ln 2}\cdot {\color{Green}\dfrac{d}{dx} 3x},\phantom{x}{\color{DarkOrange}\text{Derivative of }a^x}\text{ & }\color{Green}\text{Chain Rule}\\&= (2^{3x}\ln 2)\cdot\left({\color{DarkOrange}3\cdot \dfrac{d}{dx}x} \right ),\phantom{x}\color{DarkOrange}\text{Constant Multiple Rule}\\&= (2^{3x}\ln 2)\cdot3({\color{DarkOrange}1}),\phantom{x}\color{DarkOrange}\text{Power Rule}\\&=(3\ln 2)2^{3x} \end{aligned} | \begin{aligned}\dfrac{d}{dx} (e^{5x}) &= {\color{DarkOrange}e^{5x}}\cdot {\color{Green}\dfrac{d}{dx} 5x},\phantom{x}{\color{DarkOrange}\text{Derivative of }e^x}\text{ & }\color{Green}\text{Chain Rule}\\&= (e^{5x})\cdot\left({\color{DarkOrange}5\cdot \dfrac{d}{dx}x} \right ),\phantom{x}\color{DarkOrange}\text{Constant Multiple Rule}\\&= (e^{5x})\cdot5({\color{DarkOrange}1}),\phantom{x}\color{DarkOrange}\text{Power Rule}\\&=5e^{5x} \end{aligned} |

**Find the derivative of the following exponential functions. a. $f(x) = 3^{4x}$ b. $g(x) = 2^x – 6^x$ c. $h(x) = 5^x + e^x – 4^x$**

*Example 1*__Solution__For $f(x)$, we can apply the derivative rule for exponential function and the chain rule to differentiate it. \begin{aligned}\dfrac{d}{dx} (3^{4x}) &= {\color{DarkOrange}3^{4x} \cdot \ln 3}\cdot {\color{Green}\dfrac{d}{dx} 4x},\phantom{x}{\color{DarkOrange}\text{Derivative of }a^x}\text{ & }\color{Green}\text{Chain Rule}\\&= (3^{4x}\ln 3)\cdot\left({\color{DarkOrange}4\cdot \dfrac{d}{dx}x} \right ),\phantom{x}\color{DarkOrange}\text{Constant Multiple Rule}\\&= (3^{4x}\ln 3)\cdot4({\color{DarkOrange}1}),\phantom{x}\color{DarkOrange}\text{Power Rule}\\&=(4\ln 3)3^{4x} \end{aligned} The second function is straightforward. We’ll have to apply the difference rule and the exponential function’s derivative rule to determine $g’(x)$. \begin{aligned}\dfrac{d}{dx} (2^x – 6^x) &= \dfrac{d}{dx} 2^x – \dfrac{d}{dx} 6^x,\phantom{x}\color{DarkOrange}\text{Difference Rule}\\&= {\color{DarkOrange}(2^x\ln 2)} – {\color{DarkOrange}(6^x\ln 6)},\phantom{x}\color{DarkOrange}\text{Derivative of }a^x\end{aligned} Hence, we have $g’(x) = (2^x\ln 2) – 6^x\ln 6$. We’ll apply a similar process for $h(x)$ as shown below. \begin{aligned}\dfrac{d}{dx} (5^x +e^x -4^x) &= \dfrac{d}{dx} 5^x+ \dfrac{d}{dx} e^x – \dfrac{d}{dx}4^x,\phantom{x}\color{DarkOrange}\text{Sum & Difference Rules}\\&= \dfrac{d}{dx} 5^x- \dfrac{d}{dx}4^x+ \dfrac{d}{dx} e^x \\&= {\color{DarkOrange}(5^x \ln 5 -4^x \ln 4)} + {\color{Green}e^x},\phantom{x}{\color{DarkOrange}\text{Derivative of }a^x}\text{ & }\color{Green}\text{Derivative of }e^x\\&= 5^x \ln 5 -4^x \ln 4 +e^x \end{aligned} This shows that $h’(x) = 5^x \ln 5 -4^x \ln 4 +e^x$.

**Find the derivative of the following exponential functions. a. $f(x) = 2^{4x + 2}$ b. $g(x) = 3^{3x^2 – 4}$ c. $h(x) = e^{6^x}$**

*Example 2*__Solution__These three sets of functions are composite functions with $y = 2^x$, $y = 3^x$, and $y = e^x$, respectively. We’ll have to differentiate the inner functions for each item through chain rule. Let’s begin with $f(x) = 2^{4x + 2}$. \begin{aligned}\dfrac{d}{dx} (2^{4x + 2}) &= {\color{DarkOrange}2^{4x + 2} \cdot \ln 2}\cdot {\color{Green}\dfrac{d}{dx} (4x + 2)},\phantom{x}{\color{DarkOrange}\text{Derivative of }a^x}\text{ & }\color{Green}\text{Chain Rule}\\&= (2^{4x + 2}\ln 2)\cdot\left({\color{DarkOrange}\dfrac{d}{dx}4x + \dfrac{d}{dx} 2 } \right ),\phantom{x}\color{DarkOrange}\text{Sum Rule}\\&= (2^{4x + 2}\ln 2)\cdot\left({\color{DarkOrange}4\dfrac{d}{dx}x } + {\color{Green}0}\right ),\phantom{x}{\color{DarkOrange}\text{Constant Multiple Rule}}\text{ & }\color{Green}\text{Constant Rule} \\&= (2^{4x + 2}\ln 2)\cdot4({\color{DarkOrange}1}),\phantom{x}\color{DarkOrange}\text{Power Rule}\\&= (4\ln 2)(2^{4x + 2})\end{aligned} With the help of fundamental derivative rules, we have $f’(x) = (4\ln 2)(2^{4x + 2})$.We’ll apply a similar process to differentiate $g(x)$. \begin{aligned}\dfrac{d}{dx} (3^{3x^2 – 4}) &= {\color{DarkOrange}3^{3x^2 – 4} \cdot \ln 3}\cdot {\color{Green}\dfrac{d}{dx} (3x^2 – 4)},\phantom{x}{\color{DarkOrange}\text{Derivative of }a^x}\text{ & }\color{Green}\text{Chain Rule}\\&= (3^{3x^2 – 4} \ln 3)\cdot\left({\color{DarkOrange}\dfrac{d}{dx}3x^2 – \dfrac{d}{dx} 4 } \right ),\phantom{x}\color{DarkOrange}\text{Difference Rule}\\&= (3^{3x^2 – 4} \ln 3)\cdot\left({\color{DarkOrange}3\dfrac{d}{dx}x^2 } + {\color{Green}0}\right ),\phantom{x}{\color{DarkOrange}\text{Constant Multiple Rule}}\text{ & }\color{Green}\text{Constant Rule} \\&= (3^{3x^2 – 4} \ln 3)\cdot3({\color{DarkOrange}2x}),\phantom{x}\color{DarkOrange}\text{Power Rule}\\&= (6x\ln 3)(3^{3x^2 – 4} )\end{aligned} Hence, we have $g’(x) = (6x\ln 3)(3^{3x^2 – 4} )$. Let’s now work on $h(x)$ use the following rules to begin finding $h’(x)$.

- Derivative rule for $e^x$: $\dfrac{d}{dx} e^x = e^x$.
- Apply the chain rule and use $\dfrac{d}{dx} a^x = a^x \ln x$.

**An organism that is being studied has an initial population of $500$. After $x$ days, the population can be modeled by the function, $p(x)=500e^{0.4x}$. Jack tells his colleagues that they should expect the ratio of $p’(x)$ and $p(x)$ to be a constant. Is Jack correct?**

*Example 3*__Solution__Use the natural exponential function’s derivative rule and the chain rule to find the expression for $p’(x)$. \begin{aligned}\dfrac{d}{dx} (500e^{0.4x}) &= 500 \dfrac{d}{dx}e^{0.4x},\phantom{x}\color{DarkOrange}\text{Constant Multiple Rule}\\&= 500({\color{DarkOrange}e^{0.4x}})\cdot {\color{Green}\dfrac{d}{dx} (0.4x)},\phantom{x}{\color{DarkOrange}\text{Derivative of }e^x}\text{ & }\color{Green}\text{Chain Rule}\\&= 500e^{0.4x}\cdot\left({\color{DarkOrange}0.4\dfrac{d}{dx}x } \right ),\phantom{x}\color{DarkOrange}\text{Constant Multiple Rule}\\&= 500e^{0.4x}\cdot (0.4)({\color{DarkOrange}1}),\phantom{x}\color{DarkOrange}\text{Power Rule}\\&=200e^{0.4x}\end{aligned} Now that we have $p’(x)$, we can check whether $\dfrac{p’(x)}{p(x)}$ is indeed a constant. You can use fundamental derivative rules to further simplify the expression for $p’(x)$. \begin{aligned} \dfrac{p’(x)}{p(x)} &= \dfrac{200e^{0.4x}}{500e^{0.4x}}\\&= \dfrac{200}{500}\\&= 0.4\end{aligned} This shows that the ratio of the rate of change of the population and the actual population size is constant, so Jack is correct.

**Practice Questions**

1. Find the derivative of the following exponential functions.
a. $f(x) = 6^{2x}$
b. $g(x) = 4^x + 5^x$
c. $h(x) = 8^x – 2e^x + 6^x$
2. Find the derivative of the following exponential functions.
a. $f(x) = 8^{3x – 5}$
b. $g(x) = 6^{\sqrt{x – 1}}$
c. $h(x) = 5^{e^x}$
3. An organism that is being studied has an initial population of $800$. After $x$ days, the population can be modeled by the function, $p(x)=800e^{0.6x}$.
a. What is the ratio of $p’(x)$ and $p(x)$?
b. What is the rate of change of $p’(x)$ after $2$ days?
**Answer Key**

1.
a. $f’(x) = 9^x2^{2x + 1}\ln 6$
b. $g’(x) = 4^x\ln 4 – 5^x \ln 5$
c. $h’(x) = 8^x\ln 8 – 2e^x + 6^x \ln 6$
2.
a. $f’(x) = (3x\ln8)(8^{3x – 5})$
b. $g’(x) = \dfrac{(\ln 6)(6^{\sqrt{x – 1}})}{2\sqrt{x – 1}}$
c. $h’(x) = 5^{e^x}(e^x \ln 5)$
3.
a. $0.60$
b. $480e^{1.2} \approx 1593.67$
*Images/mathematical drawings are created with GeoGebra.*

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