23.10 Higher Derivatives
AHL 5.12
This section is for HL students. It builds on 23.9.
- higher derivatives \(\dfrac{\mathrm{d}^ny}{\mathrm{d}x^n} = f^{(n)}(x)\) and patterns in them
Higher derivatives
Higher derivatives Not in the formula booklet
Differentiating \(n\) times gives the \(n\)th derivative:
\[ \frac{\mathrm{d}^ny}{\mathrm{d}x^n} = \frac{\mathrm{d}}{\mathrm{d}x}\left(\frac{\mathrm{d}^{n-1}y}{\mathrm{d}x^{n-1}}\right) = f^{(n)}(x), \qquad n \in \mathbb{Z}^+ . \]
Note: \(\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2}\) is not the same as \(\left(\dfrac{\mathrm{d}y}{\mathrm{d}x}\right)^2\).
- \(n\): the number of times we differentiate (\(n = 1, 2, 3, \ldots\))
- \(f^{(n)}(x)\): the \(n\)th derivative of \(f\)
Example 1 Given that \(y = \tan(2x + 1)\), show that \(\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2} = 4y\dfrac{\mathrm{d}y}{\mathrm{d}x}\).
\(\dfrac{\mathrm{d}y}{\mathrm{d}x} = 2\sec^2(2x + 1) = 2\big(1 + \tan^2(2x + 1)\big) = 2(1 + y^2) = 2 + 2y^2\).
Differentiate again with respect to \(x\) (the term \(2y^2\) is differentiated implicitly):
\[ \frac{\mathrm{d}^2y}{\mathrm{d}x^2} = 4y\frac{\mathrm{d}y}{\mathrm{d}x} . \]
Sometimes the derivatives follow a pattern, and we can write a formula for \(f^{(n)}(x)\). Such formulae can be proved by mathematical induction (see 6.4).
Example 2 Let \(f(x) = \dfrac{1}{1 + x}\), \(x \neq -1\). Find \(f'(x)\), \(f''(x)\) and \(f'''(x)\), and suggest a formula for \(f^{(n)}(x)\).
\(f(x) = (1 + x)^{-1}\):
\[ f'(x) = -(1 + x)^{-2}, \qquad f''(x) = 2(1 + x)^{-3}, \qquad f'''(x) = -6(1 + x)^{-4} . \]
The signs alternate, the numbers are \(1, 2, 6 = 1!, 2!, 3!\), and the power is one more than \(n\):
\[ f^{(n)}(x) = \frac{(-1)^n\,n!}{(1 + x)^{n+1}} . \]
Exercises
1 Let \(f(x) = x\mathrm{e}^x\). Find \(f'(x)\), \(f''(x)\) and \(f'''(x)\), and suggest a formula for \(f^{(n)}(x)\).
\(f'(x) = \mathrm{e}^x + x\mathrm{e}^x = (x + 1)\mathrm{e}^x\), \(\;f''(x) = \mathrm{e}^x + (x + 1)\mathrm{e}^x = (x + 2)\mathrm{e}^x\), \(\;f'''(x) = (x + 3)\mathrm{e}^x\).
So \(f^{(n)}(x) = (x + n)\mathrm{e}^x\).
2 For \(x^2 + y^2 = 25\), show that \(\dfrac{\mathrm{d}^2y}{\mathrm{d}x^2} = -\dfrac{25}{y^3}\).
By implicit differentiation (23.8), \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = -\dfrac{x}{y}\). Quotient rule:
\[ \begin{aligned} \frac{\mathrm{d}^2y}{\mathrm{d}x^2} &= -\frac{y \cdot 1 - x\frac{\mathrm{d}y}{\mathrm{d}x}}{y^2} \\ &= -\frac{y + \frac{x^2}{y}}{y^2} \\ &= -\frac{y^2 + x^2}{y^3} \\ &= -\frac{25}{y^3} . \end{aligned} \]