Recall that the function \(f(x)=e^x\) is a special function since it is equal to the derivative: \(f^\prime(x)=e^x\). It has also been shown that \(\frac{d}{dx}\left(e^{kx}\right)=ke^{kx}\) for constants \(k\ne0\). Here, the focus is on the general exponential function \(f(x)=a^x\), where the base number \(a>0\) and \(a\ne1\). In order to find this functions derivative, some rules from exponents and logarithms will be utilized:
| Exponential Rule | Logarithm Rule | Special base \(e\) | |
|---|---|---|---|
| Inverse Rule | \(b^{\log_b{M}}=M\) | \(\log_b{b^N}=N\) | \(\begin{array}{c} e^{\ln{M}}=M \\ \ln{e^N}=N \end{array}\) |
| Change of Base | \(a^x=b^{\log_b\left(a^x\right)}\) | \(\log_bW=\large{\frac{\log_cW}{\log_cb}}\) where \(c>0\) and \(c\ne1\) |
\(\begin{array}{c} a^x=e^{\ln\left(a^x\right)} \\ \log_bW=\large{\frac{\ln W}{\ln b}} \end{array}\) |
| Using the above rules, the derivative of \(f(x)=a^x\) can be determined: | \(\begin{array}{l} f(x)=a^x \\ \implies f(x)=e^{\ln a^x} \\ \implies f(x)=e^{x\ln a},\;k=\ln a \\ \implies \frac{d}{dx}\left(f(x)\right)=\ln a\cdot e^{x\ln a} \\ \implies \frac{d}{dx}\left(f(x)\right)=\ln a\cdot e^{\ln a^x} \\ \implies \frac{d}{dx}\left(f(x)\right)=\ln a\cdot a^x\;\color{green}{\checkmark} \end{array}\) |
For an exponential function \(f(x)=a^x\), the derivative is found by: \(\large{\frac{d}{dx}}\normalsize{\left(a^x\right)=\ln a\cdot a^x}\).
Consider the logarithmic function defined by \(f(x)=\ln x\). The following table outlines the steps that derive the derivative:
| Rewrite in exponential form and take derivative of BOTH sides: | \(f(x)=\ln x\implies e^{f(x)}=x\) |
| Apply Chain Rule with inside: \(u=f(x)\) and outside: \(y=e^u\) | \(\begin{array}{l} \implies\left(\frac{dy}{du}e^{u}\right)\cdot\left(\frac{du}{dx}f(x)\right)=1 \\ \implies e^u\cdot f^\prime(x)=1 \\ \implies e^{f(x)}\cdot f^\prime(x)=1 \end{array}\) |
| Solve for \(f^\prime(x)\) and substitute back \(f(x)=\ln x\): | \(\begin{array}{l} \implies f^\prime(x)=\large{\frac{1}{e^{f(x)}}} \\ \normalsize{\implies f^\prime(x)=}\large{\frac{1}{e^{\ln x}}} \\ \normalsize{\implies f^\prime(x)=}\large{\frac{1}{x}}\;\normalsize{\color{green}{\checkmark}} \end{array}\) |
| Now by the Change of Base Rule above, the derivative of the general logarithmic function can be determined: | \(\begin{array}{l} f(x)=\log_b(x)\implies f(x)=\large{\frac{\ln x}{\ln b}} \\ \normalsize{f(x)=}\large{\frac{1}{\ln b}}\normalsize{\cdot\ln x} \\ \normalsize{f^\prime(x)=}\large{\frac{1}{\ln b}}\normalsize{\cdot}\large{\frac{1}{x}} \\ \normalsize{f^\prime(x)=}\large{\frac{1}{(\ln b)\cdot x}}\;\normalsize{\color{green}{\checkmark}} \end{array}\) |
If \(f(x)=\log_b(x)\), then \(f^\prime(x)=\large{\frac{1}{(\ln b)\cdot x}}\).
When \(b=e\), then \(f(x)=\ln x\) and \(f^\prime(x)=\large{\frac{1}{x}}\).
Straight application of formula:
$$y^\prime=\ln(5)\cdot5^x\;\color{green}{\checkmark}$$
Need the Chain Rule:
$$g^\prime(x)=\ln(3.14)\cdot3.14^{5x+2}\cdot(5)\;\color{green}{\checkmark}$$
Straight application of formula:
$$\frac{d}{dx}\left(h(x)\right)=\frac{1}{\ln(7)\cdot x}\;\color{green}{\checkmark}$$
Need the Chain Rule:
$$y^\prime=\frac{4}{\ln(3)\cdot\left(x^2-1\right)}\cdot\left(2x\right)\;\color{green}{\checkmark}$$