Calculating Higher-Order Derivatives
Introduced| English |
|---|
| second derivative/ˈsekənd dɪˈrɪvətɪv/ |
| higher-order derivatives/ˈhaɪə ˈɔːdə dɪˈrɪvətɪvz/ |
| acceleration/əkˌseləˈreɪʃn/ |
Differentiate the derivative
- Nothing stops you from differentiating $f'$ again. The result is the second derivative 二阶导数 $f''$.
- Keep going and you get higher-order derivatives 高阶导数: $f'''$, $f^{(4)}$, and so on.
- Each one measures the rate of change of the one before it.
- The second derivative is the star: it describes how the slope itself is changing.
A cubic and its slopes
y = ax³ + bx
$f'$ of a cubic is a parabola and $f''$ is a line — each derivative is one degree lower.
Just repeat the process
- Find $f'$ using any rules you need, then differentiate that to get $f''$.
- $f(x)=x^4 \Rightarrow f'(x)=4x^3 \Rightarrow f''(x)=12x^2 \Rightarrow f'''(x)=24x$.
- Each step is an ordinary derivative — no new technique, just applied again.
- Simplify $f'$ before differentiating again; it keeps the algebra clean.
For $f(x)=x^4$, what is $f'(x)$?
$f'=4x^3$, then $f''=12x^2$.
For $f(x)=x^3-2x^2+5x$, $f'(x)=6x-4$. Find $f'(2)$.
$f'(2)=6(2)-4=8$.
For $f(x)=x^3-2x^2+5x$, select all correct derivatives.
Differentiate step by step; the fourth option is a wrong simplification.
Notation for the higher orders
- Prime notation: $f''(x)$, $f'''(x)$, then $f^{(4)}(x)$ (numbers past three primes).
- Leibniz notation: $\dfrac{d^2y}{dx^2}$ for the second derivative, $\dfrac{d^3y}{dx^3}$ for the third.
- Read $\dfrac{d^2y}{dx^2}$ as "d-squared-y d-x-squared" — it means "differentiate $y$ twice."
- The two notations mean exactly the same thing.
In Leibniz notation the second derivative of $y$ is written $\dfrac{d^2 y}{dx^{\square}}$. The power $\square$ is ____.
$\dfrac{d^2y}{dx^2}$.
What the second derivative means
- $f'$ is the rate of change of $f$; $f''$ is the rate of change of $f'$.
- If $f$ is position, then $f'$ is velocity and $f''$ is acceleration — how fast the velocity changes.
- The sign of $f''$ tells you about concavity (curving up or down), coming up in Unit 5.
- So higher-order derivatives aren't busywork — $f''$ carries real, physical meaning.
The second derivative $f'$ equals $(f')^2$.
$f'$ is the derivative of $f'$, not its square.
If $f$ is position, the second derivative $f'$ represents...
$f'=$ velocity, $f''=$ acceleration.
$f''$ is the derivative of $f'$, not the square of $f'$: $f''\neq(f')^2$. And apply the chain/product/quotient rules again at each stage — differentiating $f'=\sin(x^2)$ to get $f''$ still needs the chain rule. Don't switch off the rules just because you're on the second round.
Find $f''(x)$ for $f(x)=x^3-2x^2+5x$.
- First derivative: $f'(x)=3x^2-4x+5$.
- Differentiate again: $f''(x)=6x-4$.
- (One more: $f'''(x)=6$, and $f^{(4)}(x)=0$.)
A higher-order derivative is found by differentiating repeatedly: $f''$ is the derivative of $f'$, and so on. Write them $f''(x)$ / $\frac{d^2y}{dx^2}$, etc. The second derivative is the rate of change of the first derivative (e.g. acceleration), and its sign will describe concavity. Keep applying the chain/product/quotient rules at every stage.