The second derivative and acceleration
In Year 12 Mathematical Methods (Queensland, QCAA), the second derivative \(f''(x)=\dfrac{d}{dx}\big(f'(x)\big)\) is the derivative of the first derivative — the rate of change of the gradient. Its sign gives the concavity (\(f''>0\) concave up, \(f''<0\) concave down), and for motion in a straight line the acceleration is the second derivative of displacement, \(a(t)=x''(t)\).
The second derivative of \(y=f(x)\) is what you get by differentiating twice: first \(f'(x)\), then \(f''(x)=\dfrac{d}{dx}\big(f'(x)\big)\). It is also written \(y''\) or \(\dfrac{d^{2}y}{dx^{2}}\). Because \(f'(x)\) is the gradient, \(f''(x)\) measures the rate of change of the gradient — how quickly the slope is changing.
The sign of \(f''\) describes the concavity of the graph. Where \(f''(x)>0\) the curve is concave up (curving upward, like a valley); where \(f''(x)<0\) it is concave down (curving downward, like a hill). A point of inflection is where the concavity changes, so \(f''(x)=0\) there and the sign of \(f''\) switches.
In kinematics, if \(x(t)\) is the displacement of a particle then the velocity is \(v(t)=x'(t)\) and the acceleration is \(a(t)=x''(t)\). So acceleration is the second derivative of displacement with respect to time — the rate of change of the velocity.
The second derivative is the derivative of the first derivative:
Concavity from the sign of the second derivative:
Acceleration is the second derivative of displacement:
How to work with the second derivative
- Differentiate once, then again. Find \(f'(x)\), then differentiate that to get \(f''(x)\). Apply the chain, product or quotient rule each time as needed.
- Evaluate if asked. To find \(f''(a)\), substitute \(x=a\) into the second derivative (not the original function).
- For motion, differentiate the displacement twice. Velocity \(v(t)=x'(t)\); acceleration \(a(t)=x''(t)\). Set \(a(t)=0\) to find when the acceleration is zero.
- For concavity, test the sign of \(f''\). Solve \(f''(x)>0\) for the concave-up interval and \(f''(x)<0\) for concave down; a change of sign at \(f''=0\) is a point of inflection.
Differentiate twice.
| \(f'(x)\) | \(=\) | \(6x^{2}-10x+7\) |
| \(f''(x)\) | \(=\) | \(12x-10\) |
So \(f''(x)=12x-10\).
The chain rule brings down a factor of \(3\) each time.
| \(f'(x)\) | \(=\) | \(3e^{3x}\) |
| \(f''(x)\) | \(=\) | \(9e^{3x}\) |
So \(f''(x)=9e^{3x}\).
Differentiate twice, then set \(a(t)=0\).
| \(v(t)=x'(t)\) | \(=\) | \(3t^{2}-12t+9\) |
| \(a(t)=x''(t)\) | \(=\) | \(6t-12\) |
| \(6t-12\) | \(=\) | \(0 \Rightarrow t=2\) |
The acceleration is \(a(t)=6t-12~\text{m/s}^{2}\), zero at \(t=2~\text{s}\).
The sign of \(f''\) gives the concavity.
| \(f'(x)\) | \(=\) | \(3x^{2}-6x\) |
| \(f''(x)\) | \(=\) | \(6x-6\) |
| \(6x-6\) | \(>\) | \(0 \Rightarrow x>1\) |
Concave up for \(x>1\), concave down for \(x<1\); inflection at \((1,0)\).
Common pitfalls
Frequently asked questions
What is the second derivative?
The derivative of the first derivative: differentiate once for \(f'(x)\), then again for \(f''(x)\). It is the rate of change of the gradient.
How is acceleration related to the second derivative?
For displacement \(x(t)\), velocity is \(v(t)=x'(t)\) and acceleration is \(a(t)=x''(t)\) — differentiate the displacement twice.
How does the second derivative tell you about concavity?
Its sign: \(f''>0\) is concave up, \(f''<0\) is concave down. A change of sign at \(f''=0\) is a point of inflection.
What does f''(x) = 0 mean?
A possible point of inflection (concavity may change), or in motion the instant the acceleration is zero. Confirm an inflection by checking the concavity switches.
How do you differentiate e^x, ln x or a trig function twice?
Apply the chain rule each time, e.g. \(\dfrac{d^{2}}{dx^{2}}e^{3x}=9e^{3x}\), \(\dfrac{d^{2}}{dx^{2}}\sin 2x=-4\sin 2x\), \(\dfrac{d^{2}}{dx^{2}}\ln x=-\dfrac{1}{x^{2}}\).
What is the difference between the first and second derivative?
\(f'\) is the gradient (velocity); \(f''\) is the rate of change of the gradient (acceleration). Two differentiations are needed for the second.