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Calculus AB · Unit 2 · 2.5
AP Calculus · Differentiation: Definition and Fundamental Properties
Applying the Power Rule
Bring the exponent down, reduce it by one — once you have rewritten roots and reciprocals as powers.
key idea$\frac{d}{dx}x^{n}=nx^{\,n-1}\quad\text{for EVERY real }n$
The rule
The Power Rule
\[\frac{d}{dx}\left[x^{n}\right]=n\,x^{\,n-1}\]
Bring the exponent down as a coefficient, then reduce the exponent by one. It holds for every real $n$ — positive, negative, fractional, irrational. The restriction to positive integers is a myth from early practice sets.
The two special cases worth memorising
$\dfrac{d}{dx}[x]=1$ (since $n=1$ gives $1\cdot x^{0}$) and $\dfrac{d}{dx}[c]=0$ for any constant. A constant has no $x$ to change, so its rate of change is zero.
| Written as | Rewrite as a power | Derivative |
|---|
| $x^{5}$ | — | $5x^{4}$ |
| $\sqrt{x}$ | $x^{1/2}$ | $\tfrac12 x^{-1/2}=\dfrac{1}{2\sqrt x}$ |
| $\dfrac{1}{x}$ | $x^{-1}$ | $-x^{-2}=-\dfrac{1}{x^{2}}$ |
| $\dfrac{1}{x^{3}}$ | $x^{-3}$ | $-3x^{-4}=-\dfrac{3}{x^{4}}$ |
| $\sqrt[3]{x^{2}}$ | $x^{2/3}$ | $\tfrac23 x^{-1/3}$ |
| $\dfrac{1}{\sqrt x}$ | $x^{-1/2}$ | $-\tfrac12 x^{-3/2}$ |
| $x^{\pi}$ | — | $\pi x^{\pi-1}$ |
The work is almost always in the middle column. Rewrite roots and reciprocals as powers first, and the rule applies mechanically.
Pattern recognition
rewrite first
- A root is a fractional exponent: $\sqrt[m]{x^{k}}=x^{k/m}$.
- A reciprocal is a negative exponent: $\dfrac{1}{x^{k}}=x^{-k}$.
- The exponent must be constant — $2^{x}$ is not a power function.
- The base must be the variable — $x^{x}$ is neither.
Worked examples
EX 1
Rewrite, then differentiate
$\sqrt{x}$Differentiate $f(x)=\sqrt{x}$.
1
Rewrite as a power
$\sqrt{x}=x^{1/2}$.
2
Apply the rule
$f'(x)=\tfrac12 x^{1/2-1}=\tfrac12 x^{-1/2}$.
3
Present it tidily
$f'(x)=\dfrac{1}{2\sqrt{x}}$.
4
Sanity check the domain
$f$ is defined for $x\ge0$ but $f'$ only for $x\gt 0$ — the tangent at $x=0$ is vertical.
EX 2
Negative exponent
$\dfrac{1}{x^{3}}$Differentiate $g(x)=\dfrac{1}{x^{3}}$.
1
Rewrite
$\dfrac{1}{x^{3}}=x^{-3}$.
2
Apply the rule
$g'(x)=-3x^{-3-1}=-3x^{-4}$.
3
Convert back
$g'(x)=-\dfrac{3}{x^{4}}$.
4
Check the sign
$g$ is decreasing for $x\gt 0$, so a negative derivative is right. Subtracting $1$ from $-3$ gives $-4$, not $-2$.
EX 3
A whole polynomial
$4x^{3}-\dfrac{2}{x}+7$Differentiate $h(x)=4x^{3}-\dfrac{2}{x}+7$.
1
Rewrite every term as a power
$h(x)=4x^{3}-2x^{-1}+7$.
2
Differentiate term by term
$4\cdot3x^{2}=12x^{2}$; $-2\cdot(-1)x^{-2}=2x^{-2}$; the constant $7$ gives $0$.
3
Assemble
$h'(x)=12x^{2}+2x^{-2}$.
4
Tidy
$h'(x)=12x^{2}+\dfrac{2}{x^{2}}$.
5
Watch the double negative
$-2\cdot(-1)=+2$. Two sign flips in one term is where this goes wrong.
EX 4
When the rule does not apply
$2^{x}$A student writes $\dfrac{d}{dx}\left[2^{x}\right]=x\cdot2^{\,x-1}$. What went wrong?
1
Check the form
The power rule needs a variable base and a constant exponent: $x^{n}$.
2
Identify what is actually there
$2^{x}$ has a constant base and a variable exponent — an exponential function, not a power function.
3
Name the error
The student swapped the roles. The power rule simply does not apply.
4
Give the correct result
$\dfrac{d}{dx}\left[2^{x}\right]=2^{x}\ln 2$.
5
Test yourself quickly
Ask which one is moving. If the exponent changes with $x$, it is not a power function.
Watch out
Subtract one, do not just drop a sign
For $x^{-3}$ the new exponent is $-4$, not $-2$. Reducing a negative exponent makes it more negative.
Rewrite before differentiating
$\sqrt{x}$ and $\tfrac1x$ are power functions in disguise. Trying to differentiate them in place is where most errors start.
Constant base is not a power function
$2^{x}$, $e^{x}$ and $x^{x}$ are all outside this rule. Check which of base and exponent contains the variable.
A constant differentiates to zero, not to itself
The $+7$ in Example 3 contributes nothing. It shifts the graph vertically without changing any slope.