Using the limit definition every time (The Derivative as a Function) is slow. The first shortcut covers every power of $x$, including the fractional exponents that roots become.
Key idea
$$\frac{d}{dx}x^n = nx^{n-1} \quad\text{for every real } n, \qquad \frac{d}{dx}c = 0, \qquad \frac{d}{dx}\big[c\,f(x)\big] = c\,f'(x), \qquad \frac{d}{dx}\big[f(x) \pm g(x)\big] = f'(x) \pm g'(x).$$ Bring the exponent down as a factor, then lower the exponent by one. Differentiate a sum term by term.
Why it works
For a whole number $n$, expand with the binomial theorem: $$(x + h)^n = x^n + nx^{n-1}h + (\text{terms with } h^2, h^3, \ldots)$$. Subtract $x^n$ and divide by $h$: $nx^{n-1} + (\text{terms that still contain } h)$. Let $h \to 0$ and only $nx^{n-1}$ survives. A constant has a flat graph, so its slope is 0. The sum and constant-multiple rules follow from the matching Limit Laws. The rule also holds for negative and fractional $n$; we will prove that in Logarithmic Differentiation.
Worked example: A polynomial
Differentiate $f(x) = 4x^5 - 3x^2 + 7x - 9$.
- Treat each term separately (sum rule) and keep each coefficient (constant multiple rule).
- Term by term: $4 \cdot 5x^4$, $-3 \cdot 2x$, $7 \cdot 1$, and $0$ for the constant.
Answer: $f'(x) = 20x^4 - 6x + 7$
Worked example: Roots and fractions
Differentiate $g(x) = 6\sqrt{x} - \dfrac{2}{x^3}$.
- Rewrite as powers: $g(x) = 6x^{1/2} - 2x^{-3}$.
- Power rule: $$6 \cdot \frac{1}{2}x^{-1/2} - 2 \cdot (-3)x^{-4} = 3x^{-1/2} + 6x^{-4}$$.
- Back to roots and fractions if you like: $\frac{3}{\sqrt{x}} + \frac{6}{x^4}$.
Answer: $g'(x) = \frac{3}{\sqrt{x}} + \frac{6}{x^4}$
Worked example: Simplify first
Differentiate $h(x) = \dfrac{x^2 + 1}{x}$.
- Divide each term: $h(x) = x + x^{-1}$.
- $h'(x) = 1 - x^{-2} = 1 - \frac{1}{x^2}$.
Answer: $h'(x) = 1 - \frac{1}{x^2}$
Rewrite first, then differentiate
- Roots become fractional exponents: $\sqrt[3]{x^2} = x^{2/3}$.
- Variables in a denominator become negative exponents: $\frac{5}{x^4} = 5x^{-4}$. (The 5 stays on top.)
- Products and quotients of simple terms can often be multiplied out or divided term by term first.
- After differentiating you may rewrite the answer with roots and fractions again; both forms are correct.
A quick check for any derivative: compare it with a secant slope. For $f(x) = x^3$ at $x = 2$, the rule gives $3 \cdot 4 = 12$, and $\frac{2.001^3 - 8}{0.001} = 12.006\ldots$, very close.
Common mistake
Applying the power rule to an exponential such as $2^x$ and writing $x \cdot 2^{x-1}$. The power rule is for a variable base and a constant exponent, like $x^2$. In $2^x$ the variable is in the exponent, and its derivative is $2^x \ln 2$ (see Derivatives of Exponential and Log Functions). Check: at $x = 0$, the slope of $2^x$ is about $0.69$, but $x \cdot 2^{x-1}$ gives 0.