Resources For Teachers For Tutors For Students & Parents Pricing
Create Free Account
Calculus Differentiation

Differentiation rules (power, product, quotient)

20 practice questions 0 video lessons Theory + worked examples

Differentiation Rules

California Calculus • Standard 4.4 • Differentiation

Differentiation Rules is a topic in Differentiation in the California Calculus Standards. It is aligned to Standard 4.4, which requires students to derive derivative formulas and use them to find the derivatives of algebraic functions.

The differentiation rules — the power, constant-multiple, sum, product, and quotient rules — let you find a derivative without the limit definition, for polynomials, roots, and quotients of functions.

California Calculus › Differentiation › Differentiation Rules  —  Standard 4.4

Practice 20 questions
Practice questions

Every question with a fully worked solution.

Start practising
Create a free accountTrack your progress and save your work as you go.
Create free account

Theory

The differentiation rules let you find a derivative without the limit definition. The power rule, constant-multiple and sum rules handle polynomials; the product and quotient rules handle products and fractions of functions.

Once you know a few rules, most derivatives are mechanical. The core rules:

  • Power rule: \(\dfrac{d}{dx}x^{n}=n\,x^{n-1}\) (works for negative and fractional \(n\)).
  • Constant multiple / sum: \(\dfrac{d}{dx}\big(c\,f\big)=c\,f'\) and \(\dfrac{d}{dx}(f\pm g)=f'\pm g'\).
  • Product rule: \((uv)'=u'v+uv'\).
  • Quotient rule: \(\left(\dfrac{u}{v}\right)'=\dfrac{u'v-uv'}{v^2}\).
Key idea: the derivative gives the slope of the tangent line at each \(x\). Rewrite roots and reciprocals as powers (\(\sqrt{x}=x^{1/2}\), \(\dfrac{1}{x}=x^{-1}\)) so the power rule applies.
A tangent line whose slope is given by the derivative rule The power, product, and quotient rules give a formula f prime of x for the slope of the tangent at every point. x y slope = f′(x)
The rules produce \(f'(x)\), the slope of the tangent at every point.
A function and its derivative graph The parabola f has a straight-line derivative f prime; where f falls the derivative is negative, where f rises it is positive. x y f f′
A function \(f\) (navy) and its derivative \(f'\) (teal).

The power rule and the two product-type rules:

\[\dfrac{d}{dx}x^{n}=n\,x^{n-1}\]

derivative of x to the n is n x to the n minus one

\[(uv)^{\prime}=u^{\prime}v+uv^{\prime},\qquad\left(\dfrac{u}{v}\right)^{\prime}=\dfrac{u^{\prime}v-uv^{\prime}}{v^{2}}\]

product rule u prime v plus u v prime; quotient rule u prime v minus u v prime over v squared

Quotient-rule order matters: the numerator is \(u'v-uv'\), not \(uv'-u'v\). Reversing the two terms flips the sign of the whole answer.

How to pick and apply a rule

  1. A sum of powers? Differentiate term by term with the power rule.
  2. A product of two functions? Use \((uv)'=u'v+uv'\).
  3. A fraction? Use the quotient rule, keeping the numerator order \(u'v-uv'\).
  4. Roots or reciprocals? Rewrite as powers first, then use the power rule.
Example 1 — Power and sum rules
Differentiate \(f(x)=x^4-3x^2+5\).
Solution

Differentiate term by term with the power rule.

\(f'(x)\)\(=\)\(4x^3-3\cdot 2x+0\)
\(=\)\(4x^3-6x\)

derivative is 4x cubed minus 6x

Example 2 — Product rule
Differentiate \(f(x)=(x^2+1)(x-3)\).
Solution

Use \((uv)'=u'v+uv'\) with \(u=x^2+1\), \(v=x-3\).

\(f'(x)\)\(=\)\(2x(x-3)+(x^2+1)(1)\)
\(=\)\(2x^2-6x+x^2+1\)
\(=\)\(3x^2-6x+1\)

derivative is 3x squared minus 6x plus 1

Example 3 — Quotient rule
Differentiate \(f(x)=\dfrac{x}{x^2+1}\).
Solution

Use \(\left(\dfrac{u}{v}\right)'=\dfrac{u'v-uv'}{v^2}\).

\(f'(x)\)\(=\)\(\dfrac{(1)(x^2+1)-x(2x)}{(x^2+1)^2}\)
\(=\)\(\dfrac{1-x^2}{(x^2+1)^2}\)

derivative is one minus x squared over x squared plus one squared

Example 4 — Negative and fractional powers
Differentiate \(f(x)=\sqrt{x}+\dfrac{1}{x}\).
Solution

Rewrite as powers: \(x^{1/2}+x^{-1}\), then use the power rule.

\(f'(x)\)\(=\)\(\dfrac12 x^{-1/2}-x^{-2}\)
\(=\)\(\dfrac{1}{2\sqrt{x}}-\dfrac{1}{x^2}\)

derivative uses the power rule on fractional and negative powers

Common pitfalls

Quotient-rule sign. The top is \(u'v-uv'\) — subtract in that order, or the sign of the answer flips.
The product rule is not \(f'g'\). The derivative of a product is \(u'v+uv'\), two terms, not the product of the derivatives.
Rewrite before the power rule. \(\sqrt{x}\) and \(\dfrac{1}{x^2}\) must become \(x^{1/2}\) and \(x^{-2}\) first.

Frequently asked questions

What is the power rule for derivatives?

\(\dfrac{d}{dx}x^{n}=n\,x^{n-1}\). Bring the exponent down as a factor and subtract one from it. It works for negative and fractional exponents too.

What is the product rule?

\((uv)'=u'v+uv'\): differentiate the first times the second, plus the first times the derivative of the second. It is not \(f'g'\).

What is the quotient rule?

\(\left(\dfrac{u}{v}\right)'=\dfrac{u'v-uv'}{v^{2}}\). The numerator subtracts in the order 'derivative of top times bottom minus top times derivative of bottom.'

How do you differentiate a square root?

Rewrite it as a power: \(\sqrt{x}=x^{1/2}\), then apply the power rule to get \(\dfrac{1}{2}x^{-1/2}=\dfrac{1}{2\sqrt{x}}\).

More in Differentiation
Definition of derivative
20 questions
Chain rule
20 questions
Implicit differentiation
20 questions
Derivatives of trig functions
20 questions
Derivatives of exponential and log functions
20 questions
Higher-order derivatives
20 questions