Derivative Rules
Differentiate sums, products, quotients, and compositions.
MathGPT Editorial TeamUpdated · Verify with another methodCalculus is the math of change. One half measures how fast something is changing at a single instant, and the other half adds up small amounts to give a total.
Derivative Rules: the central idea
A derivative measures instantaneous change. Rules such as product, quotient, and chain rules compress the limit definition while preserving function structure.
Words you need
- Derivative
- A derivative is the rate at which a function's output changes at one exact input, found as the limit of average slopes over shrinking intervals.
- Tangent line
- A tangent line is the straight line that touches a curve at a single point and has exactly the same slope as the curve there.
- Product rule
- The product rule says the derivative of two functions multiplied together is the first one's derivative times the second, plus the first times the second one's derivative.
- Quotient rule
- The quotient rule says the derivative of a top over a bottom is the bottom times the top's derivative minus the top times the bottom's derivative, all divided by the bottom squared.
- Chain rule
- The chain rule handles a function tucked inside another function: differentiate the outside layer as if the inside were a single letter, then multiply by the derivative of the inside.
- Higher-order derivative
- A higher-order derivative is what you get by differentiating again, so the second derivative $f''(x)$ measures how fast the rate of change is itself changing.
What to know before this lesson
Know function notation, exponent laws, slope, and how compositions such as $(3x+1)^4$ are built.
If one of those prerequisites is uncertain, use the Calculus subject guide to locate the earlier concept before memorizing a procedure.
Derivative rules: a worked example
Every step, with the arithmetic
- Step 1 - Start with the function$f(x)=3x^4-5x^2+7x-2$
- Step 2 - Power rule on the first term$3x^4 \to 4\cdot3x^{3}=12x^3$
- Step 3 - Power rule on the second term$-5x^2 \to 2\cdot(-5)x^{1}=-10x$
- Step 4 - Linear term and constant$7x \to 7$ and $-2 \to 0$, because a constant never changes
- Step 5 - Put the pieces together$f'(x)=12x^3-10x+7$
- Step 6 - Find the slope at $x=2$$f'(2)=12(8)-10(2)+7=96-20+7=83$
- Step 7 - Sanity check with a tiny secant$f(2)=40$ and $f(2.001)=40.083067$, so the average slope is $0.083067/0.001=83.067$, right next to $83$
- Step 8 - Chain rule on a nested example$\frac{d}{dx}(x^2+1)^3=3(x^2+1)^2\cdot 2x=6x(x^2+1)^2$
For $(3x+1)^4$, the power rule differentiates the outer fourth power and the chain rule multiplies by the inner derivative $3$, giving $12(3x+1)^3$.
The core derivative rules, each with a function and its finished derivative
| Rule | Function | Derivative | Worked example |
|---|---|---|---|
| Power rule | x^n | n * x^(n-1) | d/dx of x^5 = 5x^4 |
| Constant multiple | c * f(x) | c * f'(x) | d/dx of 7x^3 = 21x^2 |
| Product rule | f(x) * g(x) | f'g + f g' | d/dx of x^2 sin x = 2x sin x + x^2 cos x |
| Quotient rule | f(x) / g(x) | (f'g - f g') / g^2 | d/dx of x/(x+1) = 1/(x+1)^2 |
| Chain rule | f(g(x)) | f'(g(x)) * g'(x) | d/dx of (3x+1)^4 = 12(3x+1)^3 |
| Exponential and log | e^x and ln x | e^x and 1/x | d/dx of e^(5x) = 5e^(5x); d/dx of ln(2x) = 1/x |
The step-by-step method for derivative rules
- Identify the outermost operation: sum, product, quotient, or composition.
- Apply the matching rule without simplifying away needed factors.
- Differentiate inner functions, combine like factors, and keep the result in a form useful for the question.
How to check your answer
Compare the derivative with a numerical secant slope near a chosen point, or expand the original expression and differentiate term by term.
Compare units and qualitative behavior. A derivative should match the graph's slope, while a definite integral should agree with the signed area and scale of the interval.
A mistake that changes the mathematics
The derivative of a product is not the product of derivatives. $(fg)'=f'g+fg'$ includes two contributions.
Explain why the tempting step is invalid, then write the condition or definition that prevents it. This turns the error into a rule you can recognize in a new problem.
Where you will actually use this
Speed and acceleration
If a position function is $s(t)=3t^2$ metres after $t$ seconds, the derivative $s'(t)=6t$ gives the speed, so at $t=4$ seconds the object is moving $24$ metres per second.
Finding the best size
A company writes profit as a function of price, then sets the derivative to zero to find the price where profit stops rising and starts falling.
Medicine dosing
Drug concentration in blood rises then falls, and its derivative tells doctors how quickly the level is dropping so they know when the next dose is due.
How derivative rules connects to the rest of calculus
- Understanding limits — The derivative is defined as a limit of slopes, so limit skills are what make these rules legal shortcuts.
- Integration techniques — Substitution is the chain rule run backwards, and integration by parts is the product rule run backwards.
- Fundamental theorem of calculus — That theorem is what links these derivative rules to computing areas and totals.
Try a transfer problem
Differentiate $x^2\sin(3x)$ and label where the product rule ends and the chain rule begins.
Show the worked answer
The outermost operation in $x^2\sin(3x)$ is a multiplication, so the product rule comes first. Let $f=x^2$ and $g=\sin(3x)$. Then $f'=2x$ by the power rule. Finding $g'$ is where the chain rule begins: the outside layer is sine, whose derivative is cosine, and the inside layer is $3x$, whose derivative is $3$, so $g'=3\cos(3x)$. The product rule now assembles the answer: $f'g+fg'=2x\sin(3x)+x^2\cdot3\cos(3x)$, which is usually written $\frac{d}{dx}\left[x^2\sin(3x)\right]=2x\sin(3x)+3x^2\cos(3x)$. The product rule ends the moment you write $f'g+fg'$; the chain rule lives entirely inside the single factor $g'$. Quick check at $x=0$: the formula gives $0+0=0$, and $x^2\sin(3x)$ is flat at the origin because both $x^2$ and $\sin(3x)$ are zero there.
Work without copying the example. When finished, use the relevant focused calculator or formula reference to check the setup and result, then correct the first line where your reasoning changed. When the method feels reliable, move to calculus practice questions.
Questions about derivative rules
Is $(fg)'$ ever equal to $f'g'$?
Only by accident. Try $f=x$ and $g=x$: the real derivative of $x^2$ is $2x$, while $f'g'=1\cdot1=1$. Those agree only at $x=0.5$. The product rule $f'g+fg'$ has two terms because changing either factor changes the product.
Do I need the quotient rule, or can I always use the product rule?
You can always rewrite $f/g$ as $f\cdot g^{-1}$ and use the product rule with the chain rule, and you will get the same answer. The quotient rule is just that combination pre-simplified, so it saves algebra when the bottom is messy.
What does it mean when the derivative equals zero?
The curve is momentarily flat there. That point is a candidate for a peak, a valley, or a flat spot like the one $x^3$ has at $x=0$. Checking the sign of the derivative just before and just after the point tells you which of the three it is.
How deep can the chain rule go?
As deep as the nesting. For $\sin((3x+1)^4)$ you differentiate the sine, then the fourth power, then the $3x+1$, and multiply all three: $\cos((3x+1)^4)\cdot4(3x+1)^3\cdot3$. Each layer contributes one factor, so peel from the outside in and multiply as you go.