Power Rule | AP Calculus AB & BC Topic 2.5

1. Apply the power rule

Core concept: The power rule is a shortcut derived from the derivative definition. It differentiates powers of the independent variable, with a constant real exponent, on intervals where the power is real and differentiable. Work through polynomial, reciprocal and root examples using the supporting constant and linearity rules developed in Topic 2.6. Always keep track of the original domain.

THE POWER RULE

For a constant real exponent $n$ , the power rule gives the following derivative on $x > 0$ . Integer and suitable rational powers also allow other inputs; use the domain checks below.

\frac{d}{d x} [x^{n}] = n x^{n - 1}

Alternative Notation:

f^{'} (x) = n x^{n - 1} or \frac{d}{d x} [x^{n}] = n x^{n - 1}

In Words: Bring down the exponent as a coefficient, then subtract 1 from the original exponent.

Three-Step Process:

Step 1: Multiply by the original exponent
Step 2: Subtract 1 from the exponent
Step 3: Simplify if needed

Exponent types: The rule applies to positive, negative, fractional and irrational constant exponents on their appropriate real domains. The exponent stays constant during differentiation; a variable exponent requires another rule.

Positive integers: $x^{5}, x^{10}, x^{100}$
Negative integers: $x^{- 2}, x^{- 7}$ (from fractions like $\frac{1}{x^{2}}$ )
Fractional exponents: $x^{\frac{1}{2}}, x^{\frac{2}{3}}$ (from roots like $\sqrt{x}$ )
Irrational exponents: $x^{π}, x^{\sqrt{2}}$

Basic Power Rule Examples

Example 1: Positive Integer Exponent

Find $\frac{d}{d x} [x^{5}]$

Solution:

Original exponent: $n = 5$
Multiply by the original exponent: $5 x^{5 - 1}$
Subtract 1 from exponent: $5 - 1 = 4$
Answer: $\frac{d}{d x} [x^{5}] = 5 x^{4}$

Example 2: Another Positive Integer

If $f (x) = x^{12}$ , find $f^{'} (x)$

Solution:

Apply power rule: $f^{'} (x) = 12 x^{11}$

Answer: $f^{'} (x) = 12 x^{11}$

The exponent $n$ is constant with respect to $x$ . The output of differentiation is a rate function, not the original function. A derivative value is found by substituting an allowed input after differentiating.

2. Why the shortcut agrees with the limit definition

The rule comes from the derivative definition. For $f (x) = x^{3}$ , expand the numerator before cancelling a nonzero input change:

f^{'} (x) = lim_{h \to 0} \frac{(x + h)^{3} - x^{3}}{h} .

= lim_{h \to 0} \frac{3 x^{2} h + 3 x h^{2} + h^{3}}{h} .

= lim_{h \to 0} (3 x^{2} + 3 x h + h^{2}) = 3 x^{2} .

The quotient is undefined at $h = 0$ , but its limit exists. Cancellation is performed only for $h \neq 0$ . This proof works at every real base input, including zero.

For a positive integer exponent $n \geq 2$ , the binomial expansion gives

\frac{(x + h)^{n} - x^{n}}{h} = n x^{n - 1} + \sum_{j = 2}^{n} (\binom{n}{j}) x^{n - j} h^{j - 1} .

Every remaining term after the first contains a positive power of $h$ , so the limit is $n x^{n - 1}$ . This argument establishes positive integer powers; extensions to other real exponents need additional results. For exponent one, the difference quotient equals one directly. A constant has derivative zero.

3. Check the real domain before using a power

Positive inputs: For every fixed real exponent $n$ , $x^{n}$ is differentiable on $(0, \infty)$ with the stated rule. This includes $x^{π}$ and $x^{\sqrt{2}}$ .
Integer powers: Positive integer powers are differentiable at every real input. Negative integer powers exclude zero from both the function and derivative domains.
Rational powers: Express $r = \frac{p}{q}$ in lowest terms, with $q > 0$ . For odd $q$ , use the real-root convention $x^{\frac{p}{q}} = (\sqrt[q]{x})^{p}$ , allowing negative inputs. For even $q$ , the real domain requires nonnegative inputs. A negative exponent additionally excludes zero. At nonzero interior domain inputs the power-rule formula is valid.
Zero needs its own check: A simplified derivative formula can be undefined at zero even when the function is defined there. Decide from a limit; do not fill a hole or invent a finite derivative.

For example, $x^{\frac{4}{3}} = (\sqrt[3]{x})^{4}$ has derivative zero at zero, because its base-point quotient is $h^{\frac{1}{3}} = \sqrt[3]{h} \to 0$ . In contrast, $x^{\frac{2}{3}}$ has a cusp there. The square root has an endpoint at zero and no finite right-hand derivative there.

Do not ignore root conventions: The displayed general radical-to-power identity in the root section is stated for positive inputs. Extending it carelessly can change the function: $\sqrt{x^{2}} = | x |$ on the real line, whereas $x^{\frac{2}{2}} = x$ . These disagree for negative inputs.

Exponent zero: The constant function $f (x) = 1$ has derivative zero everywhere on its chosen domain. Do not justify its value or derivative at zero by substituting into $0 x^{- 1}$ or an undefined power.

4. Supporting rules for sums of powers

1. CONSTANT RULE

The derivative of any constant is zero.

\frac{d}{d x} [c] = 0

Examples:

$\frac{d}{d x} [5] = 0$
$\frac{d}{d x} [- 7] = 0$
$\frac{d}{d x} [π] = 0$
$\frac{d}{d x} [1000] = 0$

Why: Constants don't change, so their rate of change is zero!

2. CONSTANT MULTIPLE RULE

If $c$ is constant with respect to $x$ and $f$ is differentiable at the input, a constant multiple has derivative $c f^{'} (x)$ .

\frac{d}{d x} [c \cdot f (x)] = c \cdot \frac{d}{d x} [f (x)] = c \cdot f^{'} (x)

In words: Pull the constant out front, then take the derivative.

Examples:

$\frac{d}{d x} [3 x^{4}] = 3 \cdot \frac{d}{d x} [x^{4}] = 3 \cdot 4 x^{3} = 12 x^{3}$
$\frac{d}{d x} [- 5 x^{7}] = - 5 \cdot 7 x^{6} = - 35 x^{6}$
$\frac{d}{d x} [\frac{1}{2} x^{3}] = \frac{1}{2} \cdot 3 x^{2} = \frac{3}{2} x^{2}$

3. SUM RULE

If both functions are differentiable at the input, the derivative of their sum is the sum of their derivatives.

\frac{d}{d x} [f (x) + g (x)] = \frac{d}{d x} [f (x)] + \frac{d}{d x} [g (x)] = f^{'} (x) + g^{'} (x)

Example:

\frac{d}{d x} [x^{3} + x^{2}] = \frac{d}{d x} [x^{3}] + \frac{d}{d x} [x^{2}] = 3 x^{2} + 2 x

4. DIFFERENCE RULE

If both functions are differentiable at the input, the derivative of their difference is the difference of their derivatives.

\frac{d}{d x} [f (x) - g (x)] = \frac{d}{d x} [f (x)] - \frac{d}{d x} [g (x)] = f^{'} (x) - g^{'} (x)

Example:

\frac{d}{d x} [x^{5} - x^{3}] = 5 x^{4} - 3 x^{2}

Master Strategy: For polynomials, combine all four rules:

Step 1: Break the polynomial into individual terms (use sum/difference rules)
Step 2: For each term, pull out constants (constant multiple rule)
Step 3: Apply power rule to each $x^{n}$ term
Step 4: Constants become zero (constant rule)
Step 5: Combine all results

5. Differentiate polynomials

Example 3: Complete Polynomial

Problem: Find $\frac{d}{d x} [4 x^{3} - 7 x^{2} + 5 x - 9]$

Solution (Term-by-Term):

First term: $\frac{d}{d x} [4 x^{3}] = 4 \cdot 3 x^{2} = 12 x^{2}$
Second term: $\frac{d}{d x} [- 7 x^{2}] = - 7 \cdot 2 x = - 14 x$
Third term: $\frac{d}{d x} [5 x] = 5$ . A linear function has its constant slope at every real input, including zero.
Fourth term: $\frac{d}{d x} [- 9] = 0$ (constant disappears)
Combine: $12 x^{2} - 14 x + 5$

Answer: $12 x^{2} - 14 x + 5$

Example 4: More Complex Polynomial

Problem: If $f (x) = 2 x^{5} - 3 x^{4} + x^{3} - 6 x + 10$ , find $f^{'} (x)$

Solution:

f^{'} (x) = 2 (5 x^{4}) - 3 (4 x^{3}) + 3 x^{2} - 6 (1) + 0

f^{'} (x) = 10 x^{4} - 12 x^{3} + 3 x^{2} - 6

Answer: $f^{'} (x) = 10 x^{4} - 12 x^{3} + 3 x^{2} - 6$

6. Negative exponents and reciprocals

Rewriting with Negative Exponents

To use the power rule on fractions, rewrite using negative exponents:

\frac{1}{x^{n}} = x^{- n}

Common Conversions:

$\frac{1}{x} = x^{- 1}$
$\frac{1}{x^{2}} = x^{- 2}$
$\frac{1}{x^{3}} = x^{- 3}$
$\frac{5}{x^{4}} = 5 x^{- 4}$

Example 5: Negative Exponent

Problem: Find $\frac{d}{d x} [x^{- 3}]$

Solution:

Original exponent: $n = - 3$
Apply power rule: $- 3 \cdot x^{- 3 - 1} = - 3 x^{- 4}$
Can rewrite as: $- \frac{3}{x^{4}}$

Answer: $- 3 x^{- 4}$ or $- \frac{3}{x^{4}}$

Example 6: Fraction Form

Problem: Find $\frac{d}{d x} [\frac{2}{x^{2}} - \frac{1}{x}]$

Solution:

Rewrite with negative exponents: $2 x^{- 2} - x^{- 1}$
Apply power rule to each term:
- $\frac{d}{d x} [2 x^{- 2}] = 2 (- 2) x^{- 3} = - 4 x^{- 3}$
- $\frac{d}{d x} [- x^{- 1}] = - 1 (- 1) x^{- 2} = x^{- 2}$
Combine: $- 4 x^{- 3} + x^{- 2}$
Rewrite as fractions (optional): $- \frac{4}{x^{3}} + \frac{1}{x^{2}}$

Answer: $- 4 x^{- 3} + x^{- 2}$ or $\frac{1}{x^{2}} - \frac{4}{x^{3}}$

Examples 5 and 6 and all reciprocal conversions here exclude $x = 0$ . Rewriting a denominator as a negative exponent changes notation, not the original domain.

7. Fractional exponents and roots

Rewriting Roots as Fractional Exponents

To use the power rule on roots, convert to fractional exponents:

Radical Form	Exponent Form
$\sqrt{x}$	$x^{\frac{1}{2}}$
$\sqrt[3]{x}$	$x^{\frac{1}{3}}$
$\sqrt[n]{x}$	$x^{\frac{1}{n}}$
$\sqrt{x^{3}}$	$x^{\frac{3}{2}}$
$\sqrt[3]{x^{2}}$	$x^{\frac{2}{3}}$

General Rule:

\sqrt[n]{x^{m}} = x^{\frac{m}{n}}, x > 0, m \in ℤ, n \in ℕ, n \geq 2

Example 7: Square Root

Problem: Find $\frac{d}{d x} [\sqrt{x}]$

Solution:

Rewrite: $\sqrt{x} = x^{\frac{1}{2}}$
Apply power rule: $\frac{1}{2} x^{\frac{1}{2} - 1} = \frac{1}{2} x^{- \frac{1}{2}}$
Simplify exponent: $\frac{1}{2} - 1 = \frac{1}{2} - \frac{2}{2} = - \frac{1}{2}$
Can rewrite as: $\frac{1}{2 \sqrt{x}}$

Answer: $\frac{1}{2} x^{- \frac{1}{2}}$ or $\frac{1}{2 \sqrt{x}}$

Example 8: More Complex Root

Problem: If $f (x) = 3 \sqrt[3]{x^{2}}$ , find $f^{'} (x)$

Solution:

Rewrite: $f (x) = 3 x^{\frac{2}{3}}$
Apply constant multiple and power rule:
$f^{'} (x) = 3 \cdot \frac{2}{3} x^{\frac{2}{3} - 1} = 3 \cdot \frac{2}{3} x^{- \frac{1}{3}}$
Simplify: $f^{'} (x) = 2 x^{- \frac{1}{3}}$
Can rewrite as: $\frac{2}{\sqrt[3]{x}}$ or $\frac{2}{x^{\frac{1}{3}}}$

Answer: $f^{'} (x) = 2 x^{- \frac{1}{3}}$ or $\frac{2}{\sqrt[3]{x}}$

Example 7 domain: The square root is real for $x \geq 0$ , but its displayed derivative is valid for $x > 0$ . At zero the available quotient $\frac{\sqrt{h}}{h} = \frac{1}{\sqrt{h}}$ diverges as $h \to 0^{+}$ ; there is no finite right-hand derivative.

Example 8 domain: Interpret the cube root as the real cube root. The function is defined for all real inputs; the derivative $\frac{2}{\sqrt[3]{x}}$ is valid for $x \neq 0$ , including negative inputs. At zero its quotient is

\frac{3 \sqrt[3]{h^{2}} - 0}{h} = \frac{3}{\sqrt[3]{h}}, h \neq 0 .

The left quotient tends to $- \infty$ and the right to $+ \infty$ , giving a cusp and no finite derivative. This connects to Topic 2.4: differentiability and continuity.

8. Mixed powers and expanded products

Example 9: Everything Together

Problem: Find $\frac{d}{d x} [5 x^{4} - 3 \sqrt{x} + \frac{2}{x^{2}} + 7]$

Solution:

Rewrite everything with exponents:
$5 x^{4} - 3 x^{\frac{1}{2}} + 2 x^{- 2} + 7$
Differentiate term-by-term:
- $\frac{d}{d x} [5 x^{4}] = 20 x^{3}$
- $\frac{d}{d x} [- 3 x^{\frac{1}{2}}] = - 3 \cdot \frac{1}{2} x^{- \frac{1}{2}} = - \frac{3}{2} x^{- \frac{1}{2}}$
- $\frac{d}{d x} [2 x^{- 2}] = 2 (- 2) x^{- 3} = - 4 x^{- 3}$
- $\frac{d}{d x} [7] = 0$
Combine: $20 x^{3} - \frac{3}{2} x^{- \frac{1}{2}} - 4 x^{- 3}$

Answer: $20 x^{3} - \frac{3}{2 \sqrt{x}} - \frac{4}{x^{3}}$ or $20 x^{3} - \frac{3}{2} x^{- \frac{1}{2}} - 4 x^{- 3}$

Example 10: Expanded Form

Problem: Find $\frac{d}{d x} [x (x^{2} + 3 x - 1)]$

Solution:

Expand first: $x \cdot x^{2} + x \cdot 3 x - x \cdot 1 = x^{3} + 3 x^{2} - x$
Now differentiate:
$\frac{d}{d x} [x^{3} + 3 x^{2} - x] = 3 x^{2} + 6 x - 1$

Answer: $3 x^{2} + 6 x - 1$

Alternative: The product rule in Topic 2.8 gives the same derivative. Expanding this product first makes the power-rule calculation shorter.

Keep the domains: Example 9 and its derivative both require $x > 0$ , because the original expression contains a square root and a reciprocal. Example 10 is a polynomial after expansion and is differentiable for every real input.

9. Evaluate a derivative and write a tangent line

Reuse Example 3: $f (x) = 4 x^{3} - 7 x^{2} + 5 x - 9$ , so $f^{'} (x) = 12 x^{2} - 14 x + 5$ . The derivative is a function; a tangent slope is one of its values.

f (1) = 4 - 7 + 5 - 9 = - 7, f^{'} (1) = 12 - 14 + 5 = 3 .

The point on the curve is $(1, - 7)$ , and the tangent slope is three. Use point-slope form:

y - f (1) = f^{'} (1) (x - 1) .

y + 7 = 3 (x - 1), y = 3 x - 10 .

As an instantaneous rate, three means three output units per input unit at this input. It is not the average rate over an interval. Compare the two ideas in Topic 2.1.

A constant irrational exponent

If $p (x) = x^{π}$ for $x > 0$ , then $p^{'} (x) = π x^{π - 1}$ and $p^{'} (1) = π$ . This is a power function with a constant exponent. By contrast, $2^{x}$ has a constant base and a variable exponent, so this power-rule pattern does not apply.

10. Linear and constant special cases

Critical: Derivative of the identity function

Write $x = x^{1}$ . The exponent is one, so the derivative is the constant function one, including at zero.

\frac{d}{d x} [x] = 1

The identity function has derivative one! This is one of the most commonly forgotten facts.

Critical: Derivative of a Constant

A quantity that is constant with respect to the differentiation variable has derivative zero.

\frac{d}{d x} [c] = 0

Examples: $\frac{d}{d x} [5] = 0$ , $\frac{d}{d x} [- 100] = 0$ , $\frac{d}{d x} [π] = 0$

Important Distinction:

$\frac{d}{d x} [5 x] = 5$ (constant times x)
$\frac{d}{d x} [5] = 0$ (just a constant)

Here $x$ is the independent variable. Distinguish a constant from a constant multiplied by that variable.

11. Common derivatives and their domains

Function $f (x)$	Derivative $f^{'} (x)$	Where this derivative formula applies
$x^{n}$	$n x^{n - 1}$	$x > 0$ for arbitrary real $n$ ; extend only with valid real-power and differentiability conventions
$x$	$1$	Every real input
$c$ (constant)	$0$	Every input on the constant function's domain
$\sqrt{x} = x^{\frac{1}{2}}$	$\frac{1}{2 \sqrt{x}}$	$x > 0$
$\frac{1}{x} = x^{- 1}$	$- \frac{1}{x^{2}}$	$x \neq 0$
$\frac{1}{x^{2}} = x^{- 2}$	$- \frac{2}{x^{3}}$	$x \neq 0$
$\sqrt[3]{x} = x^{\frac{1}{3}}$	$\frac{1}{3 x^{\frac{2}{3}}}$	$x \neq 0$ ; real cube root convention
$x^{2}$	$2 x$	Every real input
$x^{3}$	$3 x^{2}$	Every real input
$x^{4}$	$4 x^{3}$	Every real input

12. Strategies, checklist and common mistakes

Essential Tips

Rewrite when useful: Converting a reciprocal or root to power notation makes the exponent visible. Preserve the real domain and any excluded inputs; this is a helpful method, not a compulsory presentation format.
Term-by-term approach: Break polynomials into separate terms and differentiate each one
Constants disappear: Any term without a variable becomes zero
Simplify exponents carefully: $\frac{1}{2} - 1 = - \frac{1}{2}$ , not $\frac{1}{1}$
Check signs: A negative exponent produces a coefficient with that sign before it is multiplied by any existing constant. Subtracting one makes the exponent more negative. For example, $\frac{d}{d x} [- x^{- 1}] = x^{- 2}$ , so the final coefficient need not be negative.
Final form flexibility: Both $x^{- 2}$ and $\frac{1}{x^{2}}$ are correct—choose what the problem asks for

Step-by-Step Checklist

Checklist for a sum of powers with constant exponents: First identify the real domain. Composite bases, variable exponents and unexpanded products may need additional rules.

REWRITE: Convert all roots and fractions to exponent form
SEPARATE: Break into individual terms (sum/difference rule)
IDENTIFY: Find constants and exponents for each term
APPLY: Use power rule: multiply by exponent, subtract 1
CONSTANTS: Pull out constant multiples, turn standalone constants to zero
SIMPLIFY: Combine like terms and simplify exponents
REFORMAT: Convert back to roots/fractions if needed

Memory Tricks

Power Rule Mnemonic:

"Drop Down, Subtract One"

Drop Down: Bring the exponent down as a coefficient
Subtract One: Reduce the exponent by 1

Four-Word Summary:

"Multiply, Drop, Subtract, Done!"

Common Mistakes to Avoid

Mistake 1: Forgetting to subtract 1 from the exponent → $\frac{d}{d x} [x^{3}] \neq 3 x^{3}$ (wrong!), it's $3 x^{2}$
Mistake 2: Confusing the derivative of $x$ with either $x$ or zero. The correct identity is $\frac{d}{d x} [x] = 1$ .
Mistake 3: Misreading a root or reciprocal exponent. For example, $\sqrt{x} = x^{\frac{1}{2}}$ and $\frac{1}{x^{2}} = x^{- 2}$ ; identify these exponents accurately whether or not you show a separate rewrite.
Mistake 4: Forgetting the constant multiple: $\frac{d}{d x} [5 x^{3}] = 5 \cdot 3 x^{2} = 15 x^{2}$ , not $3 x^{2}$
Mistake 5: Treating constants like variables. The derivative $\frac{d}{d x} [7] = 0$ , not seven.
Mistake 6: Adding instead of subtracting 1: $x^{n} \to n x^{n - 1}$ , not $n x^{n + 1}$
Mistake 7: Messing up negative exponent arithmetic: $- 2 - 1 = - 3$ , not $- 1$
Mistake 8: Confusing an unsimplified expression with an incorrect one. Both $3 \cdot \frac{1}{2} x^{2}$ and $\frac{3}{2} x^{2}$ are equivalent; combine the coefficient for clarity or when simplification is requested.

Two structure checks: Do not apply $n x^{n - 1}$ to $2^{x}$ , whose exponent varies. For $(x^{2} + 1)^{3}$ , either expand to a polynomial first or use the chain rule later; differentiating only the outer power omits an inner derivative.

Domain check after algebra: Cancelling a factor does not redefine the original function at an excluded input. For instance, $\frac{x^{2}}{x} = x$ only for $x \neq 0$ . Its derivative is one on that original domain; there is no original derivative at zero without a separately defined extension.

13. Five practice problems with solutions

Find each derivative, then check the solution and its domain.

1. A constant multiple

$f (x) = 6 x^{7}$ .

Show the power-rule solution

Multiply the coefficient by seven and subtract one from the exponent:

f^{'} (x) = 6 (7) x^{7 - 1} = 42 x^{6} .

The original function and derivative are defined for all real inputs.

2. A polynomial

$g (x) = - 4 x^{3} + 9 x^{2} - 2 x + 5$ .

Show the term-by-term solution

g^{'} (x) = - 4 (3) x^{2} + 9 (2) x - 2 + 0 .

g^{'} (x) = - 12 x^{2} + 18 x - 2 .

The constant term contributes zero. Both domains are all real inputs.

3. A root and a reciprocal

$h (x) = \sqrt{x} + \frac{3}{x}$ .

Show the rewrite and domain check

The original expression requires $x > 0$ . Rewrite it as $x^{\frac{1}{2}} + 3 x^{- 1}$ .

h^{'} (x) = \frac{1}{2} x^{- \frac{1}{2}} - 3 x^{- 2} .

h^{'} (x) = \frac{1}{2 \sqrt{x}} - \frac{3}{x^{2}}, x > 0 .

Zero is excluded by the original reciprocal; it cannot be supplied by simplification.

4. A real cube root and a negative power

$k (x) = 5 \sqrt[3]{x^{2}} - \frac{2}{x^{4}}$ .

Show the real-root solution

For $x \neq 0$ , write $k (x) = 5 x^{\frac{2}{3}} - 2 x^{- 4}$ , interpreting the fractional power with the real cube root.

k^{'} (x) = \frac{10}{3} x^{- \frac{1}{3}} + 8 x^{- 5} .

k^{'} (x) = \frac{10}{3 \sqrt[3]{x}} + \frac{8}{x^{5}}, x \neq 0 .

The minus sign on the reciprocal coefficient combines with the negative exponent to give a positive eight. The formulas apply to negative as well as positive nonzero inputs.

5. Expand before differentiating

$f (x) = (x + 1) (x - 2)$ .

Show the expansion and derivative

f (x) = x^{2} - x - 2 .

f^{'} (x) = 2 x - 1 .

Differentiate the expanded sum. The derivative of a product is not generally the product of the derivatives. This polynomial and its derivative have all real inputs as their domain.

14. Explain a derivative in an AP-style response

How to present a clear AP-style solution:

Make the method clear: A rewrite such as $\sqrt{x} = x^{\frac{1}{2}}$ is useful working. Follow the task: if it specifically asks for the limit definition, a shortcut calculation alone does not demonstrate that definition.
Follow the requested form: Simplify coefficients and exponents enough to make the answer clear. Mathematically equivalent exact forms can express the same derivative.
Use proper notation: Write $f^{'} (x)$ or $\frac{d y}{d x}$ clearly
Equivalent forms: $x^{- 2}$ and $\frac{1}{x^{2}}$ agree for nonzero inputs. Rewriting does not remove the domain restriction.
Watch for constants: Don't forget that standalone constants have derivative zero
Numerical checks: On an authorized calculator part, numerical differentiation can check a value. It does not provide an exact derivative function or establish a derivative at an excluded input. Supply the reasoning the question requests.
Communicate enough working: Identify the rule, carry out the algebra and state the result with appropriate notation and any relevant restrictions.

Common FRQ Formats:

Find $f^{'} (x)$ for a specified $f (x)$ .
"Find the derivative of..." (then simplify)
Find $\frac{d y}{d x}$ for a specified function $y = y (x)$ .
Find the tangent slope at $x = a$ : evaluate $f^{'} (a)$ where the finite derivative exists.
"Find an equation of the tangent line..." (use point-slope with derivative as slope)
Interpret an instantaneous rate at $x = a$ : evaluate the derivative and include output units per input unit.

15. Quick reference: five derivative rules

Rule Name	Formula
Power Rule	$\frac{d}{d x} [x^{n}] = n x^{n - 1}$
Constant Rule	$\frac{d}{d x} [c] = 0$
Constant Multiple	$\frac{d}{d x} [c f (x)] = c \cdot f^{'} (x)$
Sum Rule	$\frac{d}{d x} [f + g] = f^{'} + g^{'}$
Difference Rule	$\frac{d}{d x} [f - g] = f^{'} - g^{'}$

Key Conversions: $\sqrt{x} = x^{\frac{1}{2}}$ | $\frac{1}{x} = x^{- 1}$ | $\frac{d}{d x} [x] = 1$

16. Connections and next study steps

Topic 2.5 connects to:

Topic 2.6: Constant, sum, difference and constant multiple rules justify the polynomial calculations previewed here.
Topic 2.7: Derivatives of $\sin x$ , $\cos x$ , $e^{x}$ and $\ln x$ require their own formulas. In $x^{n}$ , the exponent is constant; in $e^{x}$ , the variable is in the exponent.
Topics 2.8 and 2.9: The product rule and quotient rule handle expressions that are not simply sums of powers.
Unit 3: The chain rule differentiates composite functions. For a power of an inner function, the power rule acts on the outer power and the inner derivative must also be included.
Unit 4: Motion and other contextual problems interpret derivatives of position and other quantities, with meaningful units.
Units 5 and 6: Unit 5 uses derivatives for critical points, graph behavior and optimization. Unit 6 develops integration and accumulation, including connections to differentiation.
A foundation for later work: The power rule is one important tool; choosing a rule also depends on the function's structure and domain.

Remember: For a constant real exponent, $\frac{d}{d x} [x^{n}] = n x^{n - 1}$ on the appropriate differentiability domain. Multiply the coefficient by the exponent, subtract one from that exponent, and simplify. Rewriting roots and reciprocals is often helpful. Combine powers using the supporting linearity rules, and check zero, endpoints and excluded inputs separately.

← Topic 2.4: differentiability and continuity Next: Topic 2.6 derivative rules →Review derivative definitions and notation →All Unit 2 study notes →

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