Applying the Power Rule
Introduced| English | Meaning |
|---|---|
| power/ˈpaʊə/ | The result of raising a base to an exponent, with a positive whole-number exponent representing repeated multiplication. |
| Power Rule/ˈpaʊə ruːl/ | The differentiation rule stating that the derivative of x raised to n is n multiplied by x raised to (n - 1). |
| integer/ˈɪntɪdʒə/ | A whole number that may be positive, negative or zero. |
A shortcut that replaces the limit
- Computing every derivative from the limit definition is slow. There's a pattern.
- For powers of $x$, one rule does it instantly: the Power Rule 幂法则.
-
$$\frac{d}{dx}\big[x^n\big]=n\,x^{\,n-1}$$
- "Bring the exponent down as a coefficient, then subtract one from the exponent."
A power and its tangent slope
y = x²
For $y=x^2$ the power rule gives slope $2x$ — move the point and check the tangent slope matches $2x$.
Watch it work
- $\dfrac{d}{dx}[x^3]=3x^2$ — the $3$ drops in front, the exponent goes $3\to2$.
- $\dfrac{d}{dx}[x]=1\cdot x^0=1$ — the derivative of $x$ is $1$.
- $\dfrac{d}{dx}[x^{10}]=10x^9$.
- It matches the limit definition (we found $\frac{d}{dx}[x^2]=2x$) but takes one line.
On y = x² the tangent slope at x is always 2x
What is $\dfrac{d}{dx}[x^5]$?
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Bring down $5$, reduce exponent: $5x^{5-1}=5x^4$.
If $f(x)=x^4$, find $f'(2)$.
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$f'(x)=4x^3$, so $f'(2)=4\cdot8=32$.
It works for any exponent
- $n$ can be a negative or a rational (fractional) exponent, not just a positive integer 整数.
- Negative: $\dfrac{d}{dx}[x^{-2}]=-2x^{-3}$.
- Rational: $\dfrac{d}{dx}\big[x^{1/2}\big]=\tfrac12 x^{-1/2}$.
- Same rule every time — bring down $n$, reduce the exponent by one.
$\dfrac{d}{dx}[x^{-2}]=-2x^{\square}$. The exponent $\square$ is ____.
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Reduce $-2$ by one: $-2-1=-3$.
The Power Rule differentiates which of these directly (after rewriting)?
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All are powers of $x$ after rewriting; $5^x$ is an exponential, not a power of $x$.
Rewrite first, then differentiate
- Radicals and reciprocals hide powers — rewrite them as $x^n$ before applying the rule.
- $\sqrt{x}=x^{1/2}$, so $\dfrac{d}{dx}[\sqrt x]=\tfrac12 x^{-1/2}=\dfrac{1}{2\sqrt x}$.
- $\dfrac{1}{x^3}=x^{-3}$, so $\dfrac{d}{dx}\!\left[\dfrac1{x^3}\right]=-3x^{-4}=-\dfrac{3}{x^4}$.
- Turning every term into a power is the key setup step.
Rewrite and differentiate: $\dfrac{d}{dx}[\sqrt x]$ equals...
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$\sqrt x=x^{1/2}\Rightarrow \tfrac12 x^{-1/2}=\tfrac1{2\sqrt x}$.
The Power Rule applies to $2^x$, giving $x\cdot 2^{x-1}$.
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The Power Rule needs a variable base with constant exponent; $2^x$ is an exponential (next lesson).
The Power Rule is for a variable base with a constant exponent ($x^n$). It does not apply to a constant base with a variable exponent — $\frac{d}{dx}[2^x]\neq x\,2^{x-1}$. (That is an exponential, handled in the next lesson.) And don't forget to reduce the exponent: $\frac{d}{dx}[x^{-2}]=-2x^{-3}$, not $-2x^{-2}$.
Differentiate $g(x)=\sqrt[3]{x^2}+\dfrac{1}{x}$.
- Rewrite as powers: $g(x)=x^{2/3}+x^{-1}$.
- Power rule term by term: $\tfrac23 x^{-1/3}$ and $-1\cdot x^{-2}$.
- $g'(x)=\tfrac23 x^{-1/3}-x^{-2}=\dfrac{2}{3\sqrt[3]{x}}-\dfrac1{x^2}$.
The Power Rule $\frac{d}{dx}[x^n]=n\,x^{n-1}$ works for any constant exponent — integer, negative, or rational. Rewrite radicals and reciprocals as powers first, then bring the exponent down and reduce it by one. It replaces the limit definition for power functions.