Differentiating exponentials and logarithmic functions

In a nutshell

Being able to differentiate exponentials (for example $a^{kx}$ ) and logarithms (for example $\log(kx)$ ) is a useful skill to have, since these functions (especially those involving $e$ and $\ln$ ) appear in real-life scenarios frequently.

"Natural" functions

Given the following functions for a constant $k$ :

$\begin{aligned}y&=e^{kx}\\y&=\ln(x)\end{aligned}$

their respective derivatives are given by:

$y=e^{kx}\rightarrow\boxed{\dfrac{\text dy}{\text dx}=ke^{kx}}\\\space\\y=\ln(x)\rightarrow\boxed{\dfrac{\text dy}{\text dx}=\frac1x}$

Consider now the function:

$y=\ln(kx)$

Before taking the derivative, use logarithm rules to re-express this:

$\begin{aligned}y&=\ln(kx)\\y&=\ln(k)+\ln(x)\end{aligned}$

Since $\ln(k)$ is just a constant, it follows that the derivative of $y=\ln(kx)$ is also:

$y=\ln(kx)\rightarrow\boxed{\dfrac{\text dy}{\text dx}=\frac1x}$

Example 1

Differentiate the following function with respect to $x$ :

$f(x)=4e^{6x}-2\ln(5x)$ 

Differentiating each term individually gives:

$\begin{aligned}f'(x)&=6\times4e^{6x}-\dfrac2x\\ f'(x)&=\underline{24e^{6x}-\dfrac2x}\end{aligned}$ 

General exponential

Consider now the function:

$y=a^{x}$

where $a$ is a constant. Differentiating this won't be as straightforward as differentiating $e^{x}$ since $e$ is the unique number such that differentiating $e^x$ gives itself.

To differentiate, first, using logarithm rules, rewrite this function:

$\begin{aligned}y&=e^{\ln{(a^x)}}\\y&=e^{x\ln{(a)}}\end{aligned}$

Since $\ln{a}$ is just a constant, it follows from the rule at the top that:

$\begin{aligned}\dfrac{\text dy}{\text dx}&=\ln(a)e^{x\ln(a)}\\ \dfrac{\text dy}{\text dx}&=\ln(a)e^{\ln(a^x)}\\\end{aligned}$

Thus:

$y=a^x\rightarrow\boxed{\dfrac{\text dy}{\text dx}=\ln(a)a^x}$

In a very similar way, you can show that $a^{kx}$ differentiates to $k\ln(a)a^{kx}$ .

General logarithm

Finally consider the general logarithm:

$y=\log_a{(x)}$

where $a$ is a constant. Again, since this is not the natural logarithm, it will not differentiate as easily as $\ln(x)$ does. Instead, it should first be rewritten:

$\begin{aligned}y&=\log_a(x)\\a^y&=x\\\ln(a^y)&=\ln(x)\\y\ln(a)&=\ln(x)\\y&=\frac{\ln(x)}{\ln(a)}\end{aligned}$

Since $\ln(a)$ is just a constant, it follows from the rule at the top that:

$y=\log_a(x)\rightarrow\boxed{\dfrac{\text dy}{\text dx}=\dfrac1{x\ln(a)}}$

Note: If $a=e$ , then you have the same result as before, since $\ln(e)=1$ .

In a very similar way, you can show that $\log_a(kx)$ differentiates to $\frac1{x\ln(a)}$ .

Example 2

Differentiate the following function with respect to $x$ :

$g(x)=3^{2x}+\log_{10}(4x)$ 

Differentiating each term individually gives:

$\begin{aligned}g'(x)&=2\times\ln(3)\times3^{2x}+\frac1{x\ln(10)}\\ g'(x)&=2(3^{2x})\ln(3)+\frac1{x\ln(10)}\\g'(x)&=\underline{9^x\ln(9)+\frac1{x\ln(10)}}\end{aligned}$ 