AP Calculus AB/BC · Unit 2 · Topic 2.5
Applying the Power Rule
Differentiate positive, negative, rational, and other permitted powers efficiently.
1. Topic Focus
Define derivatives as limits, estimate slopes from representations, and establish the fundamental derivative rules.
This topic: Differentiate positive, negative, rational, and other permitted powers efficiently.
2. Key Relationship
Read every symbol with its domain, direction, units, and hypotheses before applying the relationship.
3. Visual Connection
4. Worked Example
The derivative of x^(3/2) is (3/2)x^(1/2).
Write the governing relationship first, carry out the algebra cleanly, and finish with a sentence that answers the mathematical question.
5. Concept Development
A power function has the variable in the base and a constant exponent. On every interval where the power is real and differentiable, the power rule is
The exponent becomes a coefficient, and then the exponent decreases by exactly 1. This is a derivative rule, not an instruction to subtract 1 from the coefficient or from the input.
Why the pattern appears. For a positive integer \(n\), the binomial expansion begins
Substituting into the difference quotient, canceling \(x^n\), and dividing by \(h\) leaves \(nx^{n-1}\) plus terms that contain a positive power of \(h\). Those remaining terms approach 0, giving the power rule.
| Power type | Rewrite or example | Derivative | Domain reminder |
|---|---|---|---|
| Positive integer | \(x^6\) | \(6x^5\) | All real inputs |
| First power | \(x=x^1\) | \(1\) | All real inputs |
| Zero power | \(x^0=1\) | \(0\) | Treat the expression as the constant function 1 |
| Negative integer | \(x^{-3}=1/x^3\) | \(-3x^{-4}\) | Exclude \(x=0\) |
| Rational power | \(x^{3/2}=\sqrt{x^3}\) | \(\frac32x^{1/2}\) | Check where the original function is real |
Rewrite first. Radicals and reciprocals reveal their exponents when written as powers:
Then multiply by the exponent and subtract 1 using a common denominator. For example, \(3/4-1=-1/4\), not \(2/4\).
The derivative formula does not erase domain restrictions. An even root usually restricts the real domain, and a negative exponent excludes zero. Also inspect points where the differentiated formula is undefined. For \(x^{2/3}\), the function exists at 0 but the derivative formula \((2/3)x^{-1/3}\) becomes unbounded there; the original graph has a cusp and no finite derivative at 0.
Know when this rule is not enough. The basic rule above directly differentiates \(x^n\), where \(n\) is constant. A variable exponent such as \(2^x\) or \(x^x\) needs exponential or logarithmic methods. A composite power such as \((3x+1)^5\) also needs the chain rule because the base is not simply \(x\).
Quick workflow.
- Rewrite roots and reciprocals with exponents.
- Confirm that the exponent is constant and identify the real domain.
- Move the exponent in front and replace \(n\) by \(n-1\).
- Rewrite with positive exponents if that improves readability.
- State any inputs where the original function or derivative does not exist.
6. Detailed Worked Example and Error Check
Example 1: Positive integer power.
At \(x=-2\), the tangent slope is \(7(-2)^6=448\). Keep parentheses when evaluating a power at a negative input.
Example 2: Negative exponent. Rewrite before differentiating:
Both the original function and derivative exclude \(x=0\). The derivative is negative on both sides of zero.
Example 3: Even root and endpoint behavior. For \(f(x)=\sqrt{x}=x^{1/2}\),
The original function is defined at 0, but its right-hand difference quotient is \(1/\sqrt h\), which becomes unbounded as \(h\to0^+\). Thus there is no finite derivative at 0.
Example 4: Odd root with a larger numerator exponent. Let \(g(x)=\sqrt[3]{x^5}=x^{5/3}\). Then
The function is real for every \(x\), and \(g'(0)=0\). The horizontal tangent at 0 contrasts with \(x^{2/3}\), whose negative derivative exponent creates a cusp.
Example 5: Several permitted powers. After rewriting,
The combined real derivative domain is \(x>0\), because the reciprocal excludes 0 and the square-root derivative requires positive inputs.
Example 6: Context and units. If the volume of a cube is \(V(s)=s^3\) cubic centimeters, then
At side length 4 cm, volume changes at \(V'(4)=48\) cubic centimeters per centimeter of side-length change. The derivative's quotient units simplify dimensionally to square centimeters.
AP error check. Do not multiply the exponent by the base, forget to reduce the exponent by 1, change a negative exponent to positive, apply the rule directly to a variable exponent, or include points excluded from the original function.
7. AP Reasoning Routine
Identify the function structure, state the applicable rule, preserve notation and units, and check differentiability before interpreting a derivative.
- 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.
Apply the power rule and state relevant domain information.
(a) Differentiate \(f(x)=x^{11}\).
(b) Rewrite and differentiate \(g(x)=7/x^4\).
(c) Differentiate \(h(x)=\sqrt[4]{x}\), and determine whether it has a finite derivative at 0.
(d) Differentiate \(p(x)=x^{7/3}\) and evaluate \(p'(0)\).
(e) Explain why the basic power rule does not directly differentiate \(x^x\) or \((2x-1)^6\).
(f) Compare the differentiability of \(x^{2/3}\) and \(x^{4/3}\) at 0.
(g) If \(q(x)=5x^{-2}\), find \(q'(2)\).
(h) The area of a circle is \(A(r)=\pi r^2\). Find \(A'(r)\), evaluate it at \(r=3\), and give quotient units when \(r\) is measured in centimeters.
Check the solution
(a) \(f'(x)=11x^{10}\).
(b) \(g(x)=7x^{-4}\), so \(g'(x)=-28x^{-5}=-28/x^5\), with \(x\ne0\).
(c) \(h'(x)=\frac14x^{-3/4}=1/(4x^{3/4})\) for \(x>0\). The right-hand slope becomes unbounded at 0, so no finite derivative exists there.
(d) \(p'(x)=\frac73x^{4/3}\), so \(p'(0)=0\).
(e) In \(x^x\), the exponent is variable; in \((2x-1)^6\), the base is a nontrivial inner function. They require logarithmic differentiation and the chain rule, respectively.
(f) \((x^{2/3})'=(2/3)x^{-1/3}\) is unbounded with opposite signs at 0, so there is a cusp. By contrast, \((x^{4/3})'=(4/3)x^{1/3}\) exists and equals 0 at 0.
(g) \(q'(x)=-10x^{-3}\), so \(q'(2)=-10/8=-5/4\).
(h) \(A'(r)=2\pi r\), so \(A'(3)=6\pi\) square centimeters per centimeter.