Differentiating Exponentials & Logs with Other Bases
Learn to differentiate exponentials and logarithms with other bases for NSW Year 12 Mathematics Advanced (NESA outcome MAV-12-04). Functions such as \(2^x\) and \(\log_{10} x\) behave like their base-\(e\) versions, apart from a constant factor involving \(\ln a\).
You will learn that the derivative of \(a^x\) is \(a^x\) multiplied by \(\ln a\), that \(\log_a x\) has derivative \(\dfrac{1}{x\ln a}\), and how to prove both by rewriting in terms of \(e\) — so any base, such as the base 10 used in decibels and pH, can be handled with confidence.
Theory
Exponentials and logs to a base other than \(e\) pick up a factor of \(\ln a\). This Year 12 Mathematics Advanced topic (MAV-12-04) differentiates \(a^{x}\) and \(\log_a x\).
\(\dfrac{d}{dx}a^{x}=a^{x}\ln a\) and \(\dfrac{d}{dx}\log_a x=\dfrac{1}{x\ln a}\). With the chain rule, \(\dfrac{d}{dx}a^{f(x)}=f'(x)a^{f(x)}\ln a\).
When \(a=e\), \(\ln e=1\), so these collapse to \(e^{x}\) and \(\dfrac1x\).
Method
- Identify the base \(a\).
- For \(a^{x}\) multiply by \(\ln a\); for \(\log_a x\) divide by \(\ln a\).
- Apply the chain rule for a function in the exponent.
| \(\dfrac{dy}{dx}\) | \(=\) | \(2^{x}\ln2\) |
| \(\dfrac{dy}{dx}\) | \(=\) | \(\dfrac{1}{x\ln10}\) |
| \(\dfrac{dy}{dx}\) | \(=\) | \(2\cdot3^{2x}\ln3\) |
| \(\dfrac{dy}{dx}\) | \(=\) | \(5^{x}\ln5\) |
| \(\text{at }x=0\) | \(=\) | \(\ln5\) |
Common pitfalls
Frequently asked questions
What is the derivative of a to the x?
It is a to the x times ln a, where a is the base.
What is the derivative of 2 to the x?
It is 2 to the x times ln 2.
What is the derivative of log base 10 of x?
It is 1 divided by x times ln 10.
Why does base e not need the ln a factor?
Because ln e equals 1, so the factor of ln a is just 1 and disappears.