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 $$k\ln(a)a^{kx}$$.

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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:

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