AP Calculus AB/BC · Unit 3 · Topic 3.6
Calculating Higher-Order Derivatives
Differentiate repeatedly and interpret second and higher derivatives.
1. Topic Focus
Differentiate nested, implicit, inverse, and higher-order relationships by choosing procedures that match the function structure.
This topic: Differentiate repeatedly and interpret second and higher derivatives.
2. Key Relationship
Read every symbol with its domain, direction, units, and hypotheses before applying the relationship.
3. Visual Connection
4. Worked Example
For f(x)=x⁴, f′=4x³, f″=12x², and f‴=24x.
Write the governing relationship first, carry out the algebra cleanly, and finish with a sentence that answers the mathematical question.
5. Concept Development
A derivative is itself a function, so it can be differentiated again. The derivative of \(f'\) is the second derivative, the derivative of \(f''\) is the third derivative, and the process can continue whenever each new derivative exists.
| Order | Prime notation | Leibniz notation | Meaning |
|---|---|---|---|
| First | \(f'(x)\) | \(dy/dx\) | Rate of change of \(f\) |
| Second | \(f''(x)\) | \(d^2y/dx^2\) | Rate of change of \(f'\) |
| Third | \(f'''(x)\) | \(d^3y/dx^3\) | Rate of change of \(f''\) |
| \(n\)th | \(f^{(n)}(x)\) | \(d^ny/dx^n\) | Derivative applied \(n\) times |
The parentheses in \(f^{(n)}\) indicate derivative order, not an exponent. Likewise, \(d^2y/dx^2\) is not \((dy/dx)^2\); it means \(d/dx\) applied to \(dy/dx\).
Repeated-differentiation routine.
- Write each derivative on its own line and label its order.
- Simplify enough to make the next differentiation reliable.
- Reassess the expression's structure; product, quotient, or chain rules may be needed again.
- Continue only to the requested order.
- Check patterns, domains, signs, and units before evaluating at a point.
Polynomial pattern. For an integer \(m\ge0\), repeated use of the power rule gives
The \(m\)th derivative is \(m!\), and every derivative of order greater than \(m\) is zero. For a polynomial, the degree drops by one with each derivative until only zero remains.
Patterns that do not terminate.
Sine and cosine repeat every four derivative orders. The coefficient \(a^n\) records one chain-rule factor of \(a\) at each differentiation.
Meaning from derivative graphs. The value \(f''(a)\) is the slope of the graph of \(f'\) at \(x=a\), and \(f'''(a)\) is the slope of \(f''\). Thus \(f''(a)>0\) means \(f'\) has positive slope at that point. Over an interval where \(f''>0\), the function \(f'\) is increasing; over an interval where \(f'''<0\), the function \(f''\) is decreasing.
Motion and units. If position \(s(t)\) is measured in meters and time in seconds, then
Velocity has units meters per second, acceleration meters per second squared, and jerk meters per second cubed. Each differentiation divides the units by one additional input unit.
Implicit higher derivatives. When \(y\) is defined implicitly, differentiate the entire first-derivative equation again. Every occurrence of \(y\) still depends on \(x\). For example, \(d(yy')/dx=(y')^2+yy''\), not simply \(yy''\).
6. Detailed Worked Example and Error Check
Example 1: Polynomial derivatives terminate. Let \(f(x)=x^5-3x^3+2x\). Then
Example 2: Exponential pattern. For \(g(x)=e^{3x}\), each derivative supplies another factor of 3:
In particular, \(g^{(5)}(0)=3^5=243\).
Example 3: Trigonometric cycle. For \(h(x)=\cos(2x)\),
After four differentiations the original function returns with a factor of \(2^4\). Hence \(h^{(7)}(x)=128\sin(2x)\).
Example 4: Product rule at every stage. Let \(p(x)=x^2e^x\). Then
Factoring out \(e^x\) after each step makes the next product-rule calculation easier to audit.
Example 5: Second derivative of an implicit curve. For \(x^2+y^2=25\), the first derivative equation is
Differentiate this equation again:
Using \(y'=-x/y\) and \(x^2+y^2=25\),
Example 6: Position, velocity, acceleration, and jerk. Suppose
Then
At \(t=2\), \(v=-8\), \(a=4\), and \(j=24\). These values describe different rates and must carry different units.
Example 7: Recognize a general logarithmic pattern. Starting with \(q(x)=\ln x\),
The coefficient is factorial and the signs alternate:
AP error check. Do not interpret \(f^{(4)}\) as \(f^4\), square the first derivative instead of finding the second derivative, stop applying product or chain rules after the first step, lose repeated inner factors, confuse velocity with acceleration, or ignore that higher derivatives can have smaller domains.
7. AP Reasoning Routine
Mark inner and outer functions, track every derivative factor, solve algebraically for the requested derivative, and verify the result's domain.
- Identify the representation and requested quantity.
- State the rule or theorem and verify its conditions.
- Keep exact values until the final requested approximation.
- Interpret sign, units, interval, and context.
Show enough intermediate derivatives to make the pattern or procedure clear.
(a) Find the first four derivatives of \(f(x)=2x^5-x^3+7\).
(b) Find \(d^8(\sin x)/dx^8\).
(c) If \(g(x)=e^{-2x}\), find \(g^{(5)}(x)\).
(d) Find \(p''(x)\) for \(p(x)=xe^x\).
(e) If \(x^2+4y^2=16\), find \(d^2y/dx^2\) in terms of \(y\).
(f) A particle has position \(s(t)=t^3-6t^2+9t\). Find velocity, acceleration, and jerk, then evaluate all three at \(t=2\).
(g) If \(A(t)\) is measured in liters and \(t\) in minutes, state the units of \(A'''(t)\).
(h) Suppose \(r(x)=f'(x)\), \(r'(3)=5\), and \(r''(3)=-2\). Find \(f''(3)\) and \(f'''(3)\).
Check the solution
(a) \(f'=10x^4-3x^2\), \(f''=40x^3-6x\), \(f'''=120x^2-6\), and \(f^{(4)}=240x\).
(b) Sine derivatives repeat every four orders. Since \(8\equiv0\pmod4\), \(d^8(\sin x)/dx^8=\sin x\).
(c) \(g^{(n)}=(-2)^ne^{-2x}\), so \(g^{(5)}=-32e^{-2x}\).
(d) \(p'=e^x(x+1)\), so \(p''=e^x(x+2)\).
(e) First, \(2x+8yy'=0\). Differentiating again gives \(2+8[(y')^2+yy'']=0\). Using \(y'=-x/(4y)\) and \(x^2+4y^2=16\) yields \(y''=-1/y^3\).
(f) \(v=3t^2-12t+9\), \(a=6t-12\), and \(j=6\). At \(t=2\), \(v=-3\), \(a=0\), and \(j=6\).
(g) Liters per minute cubed, written \(\text{L}/\text{min}^3\).
(h) Since \(r=f'\), \(r'=f''\) and \(r''=f'''\). Therefore \(f''(3)=5\) and \(f'''(3)=-2\).