23.7 Derivatives of aˣ and logₐ x
AHL 5.15
This section is for HL students. It builds on 23.6.
- the derivative of \(a^x\) for any base \(a > 0\)
- the derivative of \(\log_a x\)
Exponential functions with any base
For \(a > 0\), we can write \(a\) as a power of \(\mathrm{e}\): \(\;a = \mathrm{e}^{\ln a}\). So
\[ a^x = \left(\mathrm{e}^{\ln a}\right)^x = \mathrm{e}^{(\ln a)x} . \]
Now \(\ln a\) is just a constant, so by the chain rule
\[ \frac{\mathrm{d}}{\mathrm{d}x}\left(a^x\right) = (\ln a)\,\mathrm{e}^{(\ln a)x} = a^x\ln a . \]
Derivatives of \(a^x\) and \(a^{f(x)}\) (\(a > 0\)) Partly in the HL formula booklet
\[ f(x) = a^x \;\Rightarrow\; f'(x) = a^x(\ln a) \]
In the HL booklet
More generally, by the chain rule,
\[ \frac{\mathrm{d}}{\mathrm{d}x}\left(a^{f(x)}\right) = f'(x)\,a^{f(x)}\ln a . \]
- \(a\): the base, a positive constant
- \(f(x)\): a function of \(x\) (the power)
- \(\ln a\): the natural logarithm of \(a\)
Example 1 Find \(\dfrac{\mathrm{d}y}{\mathrm{d}x}\) when
(a) \(y = 2^{x^2}\)
(b) \(y = 7^{1 - 4x}\).
(a) \(\;\dfrac{\mathrm{d}y}{\mathrm{d}x} = 2x \cdot 2^{x^2}\ln 2\).
(b) \(\;\dfrac{\mathrm{d}y}{\mathrm{d}x} = -4 \cdot 7^{1 - 4x}\ln 7\).
Logarithms with any base
For a logarithm with another base, change the base first: \(\log_a x = \dfrac{\ln x}{\ln a}\) (see 1.4). Since \(\dfrac{1}{\ln a}\) is a constant:
Derivative of \(\log_a x\) In the HL formula booklet
\[ f(x) = \log_a x \;\Rightarrow\; f'(x) = \frac{1}{x\ln a} \]
- \(a\): the base of the logarithm (\(a > 0\), \(a \neq 1\))
- \(x\): a positive number (\(x > 0\))
Example 2 Differentiate \(y = \log_2(x^2 + 1)\) with respect to \(x\).
\(y = \dfrac{\ln(x^2 + 1)}{\ln 2}\), so \(\;\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{1}{\ln 2} \times \dfrac{2x}{x^2 + 1} = \dfrac{2x}{(x^2 + 1)\ln 2}\).
Common errors
| ✗ Wrong | ✓ Right | Why |
|---|---|---|
| \(\dfrac{\mathrm{d}}{\mathrm{d}x}\left(3^x\right) = x \cdot 3^{x - 1}\) | \(3^x\ln 3\) | The variable is in the exponent. |
Exercises
1 Differentiate \(y = 3^x + \log_3 x\).
\(\dfrac{\mathrm{d}y}{\mathrm{d}x} = 3^x\ln 3 + \dfrac{1}{x\ln 3}\).
2 Let \(f(x) = x \cdot 2^x\). Find \(f'(x)\), and find the exact value of \(x\) for which \(f'(x) = 0\).
Product rule:
\[ f'(x) = 1 \cdot 2^x + x \cdot 2^x\ln 2 = 2^x(1 + x\ln 2) . \]
\(2^x > 0\) for all \(x\), so \(f'(x) = 0\) only when \(1 + x\ln 2 = 0\), i.e. \(x = -\dfrac{1}{\ln 2}\).