The Central Idea
Derivative rules follow the structure of the function being differentiated. Three especially important structures are:
- a product of changing factors;
- a quotient of changing expressions;
- a composition, where one function is inside another.
Each structure has its own rule:
\[ \boxed{(fg)'=f'g+fg'} \] \[ \boxed{\left(\frac{f}{g}\right)'=\frac{f'g-fg'}{g^2}} \] \[ \boxed{(f\circ g)'(x)=f'(g(x))g'(x)}. \]The practical principle is simple: identify the outermost structure first, apply its rule, and then differentiate the pieces that remain.
Structure Comes Before Calculation
Compare the following functions:
\[ x^2+x^5, \] \[ x^2(x^5+1), \] \[ \frac{x^2}{x^5+1}, \]and
\[ (x^2+1)^5. \]They contain similar symbols, but their outer structures are different. The first is a sum, the second is a product, the third is a quotient, and the fourth is a composition. Their derivatives therefore require different workflows.
The Product Rule
Suppose
\[ y=f(x)g(x), \]where both factors depend on \(x\). Then
\[ \boxed{\frac{d}{dx}[f(x)g(x)]=f'(x)g(x)+f(x)g'(x)}. \]In compact notation,
\[ \boxed{(fg)'=f'g+fg'}. \]One factor changes while the other is temporarily held in place, and then the roles reverse. The two resulting contributions are added.
The Derivative of a Product Is Not the Product of the Derivatives
A common mistake is to write
\[ (fg)'=f'g'. \]This is generally false.
For example, let
\[ f(x)=x^2, \qquad g(x)=x^3. \]Their product is
\[ f(x)g(x)=x^5, \]so the actual derivative is
\[ 5x^4. \]But multiplying the individual derivatives would give
\[ (2x)(3x^2)=6x^3, \]which is different. The product rule is necessary because both factors are changing at the same time.
Why the Product Rule Works
Let
\[ F(x)=f(x)g(x). \]Its difference quotient is
\[ \frac{f(x+h)g(x+h)-f(x)g(x)}{h}. \]Add and subtract \(f(x)g(x+h)\) in the numerator:
\[ \frac{f(x+h)g(x+h)-f(x)g(x+h)+f(x)g(x+h)-f(x)g(x)}{h}. \]Group the terms:
\[ g(x+h)\frac{f(x+h)-f(x)}{h} + f(x)\frac{g(x+h)-g(x)}{h}. \]As \(h\to0\), differentiability implies continuity, so \(g(x+h)\to g(x)\). The two difference quotients approach \(f'(x)\) and \(g'(x)\). Therefore,
\[ \boxed{F'(x)=f'(x)g(x)+f(x)g'(x)}. \]Product Rule Example
Differentiate
\[ y=x^2(x^3+4). \]Let
\[ f(x)=x^2, \qquad g(x)=x^3+4. \]Then
\[ f'(x)=2x, \qquad g'(x)=3x^2. \]Apply the product rule:
\[ y'=(2x)(x^3+4)+x^2(3x^2). \]Simplifying,
\[ \boxed{y'=5x^4+8x}. \]Sometimes Expanding First Is Easier
The previous function can also be expanded before differentiating:
\[ x^2(x^3+4)=x^5+4x^2. \]Then
\[ y'=5x^4+8x. \]Both methods are valid. If expansion turns a product into a short sum of powers, simplifying first may be faster. The product rule becomes more valuable when expansion is inconvenient or when the factors should remain structurally visible.
Product Rule with More Than Two Factors
The product rule can be extended to several factors. For three functions,
\[ y=fgh, \]differentiate one factor at a time while leaving the others unchanged:
\[ \boxed{y'=f'gh+fg'h+fgh'}. \]The same idea continues for larger products: each term in the derivative contains one differentiated factor.
The Quotient Rule
Suppose
\[ y=\frac{f(x)}{g(x)}, \qquad g(x)\ne0. \]Then
\[ \boxed{ y'=\frac{f'(x)g(x)-f(x)g'(x)}{[g(x)]^2} }. \]The order in the numerator matters. The derivative of the numerator times the original denominator comes first, followed by subtraction of the original numerator times the derivative of the denominator.
A Quotient-Rule Memory Pattern
If the numerator is called the high expression and the denominator the low expression, a common memory pattern is:
\[ \boxed{ \frac{\text{low}\cdot d(\text{high})-\text{high}\cdot d(\text{low})} {(\text{low})^2} }. \]The mnemonic is useful only if the subtraction order is preserved. Writing the formula symbolically before substituting expressions is often safer.
Why the Quotient Rule Works
Let
\[ y=\frac{f}{g}. \]Then
\[ yg=f. \]Differentiate both sides using the product rule:
\[ y'g+yg'=f'. \]Solve for \(y'\):
\[ y'g=f'-yg'. \]Since \(y=f/g\),
\[ y' =\frac{f'-\frac{f}{g}g'}{g} =\frac{f'g-fg'}{g^2}. \]Thus,
\[ \boxed{\left(\frac{f}{g}\right)'=\frac{f'g-fg'}{g^2}}. \]Quotient Rule Example
Differentiate
\[ y=\frac{x^2+1}{x-3}. \]Let
\[ f(x)=x^2+1, \qquad g(x)=x-3. \]Then
\[ f'(x)=2x, \qquad g'(x)=1. \]Apply the quotient rule:
\[ y'=\frac{(2x)(x-3)-(x^2+1)(1)}{(x-3)^2}. \]Simplify the numerator:
\[ 2x^2-6x-x^2-1=x^2-6x-1. \]Therefore,
\[ \boxed{y'=\frac{x^2-6x-1}{(x-3)^2}}, \qquad x\ne3. \]Simplify a Quotient Before Differentiating When Possible
Not every fraction needs the quotient rule. Consider
\[ f(x)=\frac{x^4+3x^2}{x^2}, \qquad x\ne0. \]Simplify first:
\[ f(x)=x^2+3. \]Then
\[ \boxed{f'(x)=2x}, \qquad x\ne0. \]The derivative formula is simple, but the original function was not defined at \(x=0\). Algebraic simplification does not erase the original domain restriction.
The Chain Rule
The chain rule differentiates composite functions. If
\[ y=f(g(x)), \]then
\[ \boxed{\frac{dy}{dx}=f'(g(x))g'(x)}. \]Differentiate the outer function while leaving the inner function in place, then multiply by the derivative of the inner function.
Recognizing a Composite Function
A composite function has one function inserted into another. For example,
\[ (3x+1)^5 \]can be viewed as
\[ u^5 \]with
\[ u=3x+1. \]The fifth power is the outer operation. The linear expression \(3x+1\) is the inner function.
Chain Rule Example: A Power of a Linear Expression
Differentiate
\[ y=(3x+1)^5. \]Differentiate the outer fifth power:
\[ 5(3x+1)^4. \]Differentiate the inner expression:
\[ \frac{d}{dx}(3x+1)=3. \]Multiply the two pieces:
\[ \boxed{y'=15(3x+1)^4}. \]Why the Inner Derivative Matters
Compare
\[ x^5 \]with
\[ (3x+1)^5. \]The derivative of \(x^5\) is
\[ 5x^4. \]But the inner quantity \(3x+1\) changes three times as fast as \(x\), so the composite derivative acquires an additional factor of \(3\):
\[ \frac{d}{dx}[(3x+1)^5] =5(3x+1)^4(3). \]The chain rule accounts for how quickly the input to the outer function is itself changing.
Chain Rule in Leibniz Notation
Suppose
\[ y=f(u) \]and
\[ u=g(x). \]Then the chain rule may be written suggestively as
\[ \boxed{\frac{dy}{dx}=\frac{dy}{du}\frac{du}{dx}}. \]This notation helps track intermediate variables. The symbols resemble fractions and provide a useful cancellation pattern, but derivatives are defined through limits rather than as ordinary fractions.
Chain Rule Example: A Radical
Differentiate
\[ f(x)=\sqrt{2x^2+3}. \]Rewrite the radical as a power:
\[ f(x)=(2x^2+3)^{1/2}. \]Differentiate the outer power and multiply by the derivative of the inner expression:
\[ f'(x) =\frac12(2x^2+3)^{-1/2}(4x). \]Therefore,
\[ \boxed{f'(x)=\frac{2x}{\sqrt{2x^2+3}}}. \]Chain Rule Example: A Negative Power
Differentiate
\[ f(x)=(x^2+4)^{-2}. \]Differentiate the outer power:
\[ -2(x^2+4)^{-3}. \]Multiply by the derivative of \(x^2+4\):
\[ f'(x)=-2(x^2+4)^{-3}(2x). \]Thus,
\[ \boxed{f'(x)=-\frac{4x}{(x^2+4)^3}}. \]Nested Chain Rules
A composition can contain more than two layers. Suppose
\[ y=[(2x+1)^3+4]^5. \]Work from the outside inward. Differentiate the fifth power:
\[ 5[(2x+1)^3+4]^4. \]Then differentiate the next layer:
\[ 3(2x+1)^2. \]Finally differentiate the innermost expression:
\[ 2. \]Multiply all three contributions:
\[ \boxed{ y'=30[(2x+1)^3+4]^4(2x+1)^2 }. \]The Generalized Power Rule
The ordinary power rule and chain rule combine into an especially useful pattern. If \(u=u(x)\), then
\[ \boxed{\frac{d}{dx}[u^n]=nu^{n-1}u'}. \]In function notation,
\[ \boxed{\frac{d}{dx}[(g(x))^n] =n[g(x)]^{n-1}g'(x)}. \]This is one of the most frequently used chain-rule patterns.
Combining the Product and Chain Rules
Differentiate
\[ y=x^2(x^3+1)^4. \]The outermost structure is a product, so begin with the product rule:
\[ y'=2x(x^3+1)^4 +x^2\frac{d}{dx}[(x^3+1)^4]. \]The second factor requires the chain rule:
\[ \frac{d}{dx}[(x^3+1)^4] =4(x^3+1)^3(3x^2). \]Therefore,
\[ \boxed{ y'=2x(x^3+1)^4+12x^4(x^3+1)^3 }. \]Factoring a Derivative Can Reveal Structure
The derivative from the previous example can be factored:
\[ y'=2x(x^3+1)^4+12x^4(x^3+1)^3. \]The common factor is
\[ 2x(x^3+1)^3. \]Thus,
\[ \boxed{ y'=2x(x^3+1)^3(7x^3+1) }. \]Expanded and factored derivative forms are equivalent. The factored form may be more useful later when studying zeros, critical points, and signs.
Combining the Quotient and Chain Rules
Differentiate
\[ y=\frac{(x^2+1)^3}{x-2}. \]The outermost structure is a quotient. Let
\[ f(x)=(x^2+1)^3, \qquad g(x)=x-2. \]The numerator derivative requires the chain rule:
\[ f'(x)=3(x^2+1)^2(2x)=6x(x^2+1)^2. \]Also,
\[ g'(x)=1. \]Apply the quotient rule:
\[ \boxed{ y'= \frac{6x(x^2+1)^2(x-2)-(x^2+1)^3}{(x-2)^2} }. \]It is usually safer to apply the rules correctly before deciding how much algebraic simplification is useful.
One Function Can Require Several Rules
A derivative problem does not always have one rule. Consider
\[ y=\frac{x^2(3x+1)^4}{x^3+2}. \]Its outermost structure is a quotient. The numerator is a product. One factor in that product is a composition. A complete derivative therefore requires:
- the quotient rule;
- the product rule inside the numerator derivative;
- the chain rule inside the derivative of \((3x+1)^4\);
- the ordinary power rule for the remaining powers of \(x\).
Complex differentiation is usually a matter of applying familiar rules in the correct hierarchy.
Work from the Outside In
For a complicated expression, identify its outermost operation before doing any differentiation.
If the function looks like
\[ A(x)B(x), \]begin with the product rule. If it looks like
\[ \frac{A(x)}{B(x)}, \]begin with the quotient rule. If it looks like
\[ F(G(x)), \]begin with the chain rule. Then inspect \(A\), \(B\), or \(G\) to see whether additional rules are needed inside.
Parentheses Often Reveal the Chain Rule
Expressions such as
\[ (x^2+5)^7, \qquad \sqrt{4x^3-1}, \qquad \frac{1}{(2x+3)^4} \]all contain an inner expression being acted on by an outer operation. Rewriting powers and radicals can make that structure especially clear:
\[ \sqrt{4x^3-1}=(4x^3-1)^{1/2} \]and
\[ \frac{1}{(2x+3)^4}=(2x+3)^{-4}. \]Once the composition is visible, the generalized power rule can be applied immediately.
Do Not Use the Quotient Rule When a Negative Power Is Simpler
Consider
\[ f(x)=\frac{1}{(x^2+1)^3}. \]Rather than treating this as a quotient, rewrite it as
\[ f(x)=(x^2+1)^{-3}. \]Then the chain rule gives
\[ f'(x)=-3(x^2+1)^{-4}(2x). \]Therefore,
\[ \boxed{f'(x)=-\frac{6x}{(x^2+1)^4}}. \]Choosing a simpler equivalent representation can eliminate unnecessary work.
Derivative Rules Preserve Meaning
These formulas are computational tools, but the derivative still represents local rate of change. If
\[ A(x)=f(x)g(x), \]then the product rule describes how the total product changes because changes in either factor affect the result. If
\[ y=f(g(x)), \]then the chain rule combines the outer response rate with the rate at which the inner input is changing. The rules compress rate-of-change reasoning into symbolic form.
A Simple Rate Interpretation of the Product Rule
Suppose a rectangle has width \(w(t)\) and height \(h(t)\). Its area is
\[ A(t)=w(t)h(t). \]Differentiate with respect to time:
\[ \boxed{ \frac{dA}{dt} =\frac{dw}{dt}h+w\frac{dh}{dt} }. \]Area can change because the width changes, because the height changes, or because both change. The product rule records both first-order contributions.
A Simple Rate Interpretation of the Chain Rule
Suppose a quantity \(y\) depends on \(u\), while \(u\) depends on time \(t\):
\[ y=f(u), \qquad u=g(t). \]Then
\[ \boxed{ \frac{dy}{dt}=\frac{dy}{du}\frac{du}{dt} }. \]The rate of change of \(y\) with respect to time is the response of \(y\) to a change in \(u\), multiplied by the rate at which \(u\) itself changes with time.
Differentiability Still Matters
Derivative rules apply where the required component derivatives exist and where the original expression is defined. For a quotient,
\[ \frac{f(x)}{g(x)}, \]the denominator must satisfy
\[ g(x)\ne0. \]A symbolic derivative formula should always be interpreted together with the domain and differentiability conditions of the original function.
Common Error: Multiplying the Derivatives
If
\[ y=f(x)g(x), \]do not write
\[ y'=f'(x)g'(x). \]The correct product rule is
\[ \boxed{y'=f'g+fg'}. \]Common Error: Reversing the Quotient-Rule Subtraction
The quotient rule is
\[ \boxed{ \left(\frac{f}{g}\right)'=\frac{f'g-fg'}{g^2} }. \]Reversing the numerator to \(fg'-f'g\) changes the sign of the derivative. Write the formula before substituting if the order is easy to lose.
Common Error: Forgetting to Square the Denominator
If
\[ y=\frac{f}{g}, \]the denominator of the derivative is
\[ \boxed{g^2}, \]not merely \(g\). Parentheses are important when the denominator is an expression:
\[ [g(x)]^2. \]Common Error: Forgetting the Inner Derivative
For
\[ y=(5x-2)^6, \]writing
\[ 6(5x-2)^5 \]differentiates only the outer layer. The inner derivative is \(5\), so
\[ \boxed{y'=30(5x-2)^5}. \]Common Error: Using the Chain Rule on a Product
The expression
\[ x^2(x+1)^3 \]has a product as its outermost structure. The factor \((x+1)^3\) contains a composition, but the first rule must be the product rule. Rule selection follows the outside structure first.
Common Error: Differentiating the Denominator Alone
For a quotient such as
\[ \frac{x^2+1}{x+4}, \]the derivative is not obtained by differentiating the numerator and denominator separately. In general,
\[ \left(\frac{f}{g}\right)' \ne \frac{f'}{g'}. \]The quotient rule accounts for the simultaneous change of both numerator and denominator.
A Reliable Rule-Selection Procedure
- Simplify obvious algebra when doing so clearly reduces the work.
- Identify the outermost operation of the function.
- Use the sum or difference rule if the outermost structure is addition or subtraction.
- Use the product rule if changing factors are multiplied.
- Use the quotient rule if one changing expression is divided by another and simplification is not easier.
- Use the chain rule when one function is applied to another.
- Differentiate each internal piece using whatever rules it requires.
- Preserve parentheses until the rule structure is secure.
- Simplify the derivative only after the differentiation is correct.
- Check domain restrictions and, when useful, verify by an alternate form.
Example: Full Mixed-Rule Workflow
Differentiate
\[ y=\frac{x(2x+1)^3}{x^2+4}. \]The outermost structure is a quotient. Define
\[ N(x)=x(2x+1)^3, \qquad D(x)=x^2+4. \]Differentiate the denominator:
\[ D'(x)=2x. \]The numerator is a product:
\[ N'(x) =(1)(2x+1)^3 +x\left[3(2x+1)^2(2)\right]. \]Therefore,
\[ N'(x)=(2x+1)^3+6x(2x+1)^2. \]Apply the quotient rule:
\[ \boxed{ y'= \frac{ \left[(2x+1)^3+6x(2x+1)^2\right](x^2+4) -x(2x+1)^3(2x) }{(x^2+4)^2} }. \]A fully expanded answer is not required unless a particular application benefits from expansion.
Checking a Derivative by Rewriting
When a function has two useful algebraic forms, differentiating both forms can provide a strong check. For example,
\[ y=x(x+2) \]gives by the product rule
\[ y'=(1)(x+2)+x(1)=2x+2. \]Expanding first gives
\[ y=x^2+2x, \]so
\[ y'=2x+2. \]Agreement between the two methods supports the calculation.
Practice 1: Product Rule
Differentiate:
\[ y=x^3(x^2+5). \]Practice 2: Quotient Rule
Differentiate:
\[ y=\frac{x^2-4}{x+1}. \]Practice 3: Chain Rule
Differentiate:
\[ y=(4x-3)^7. \]Practice 4: Generalized Power Rule
Differentiate:
\[ f(x)=(x^3+2x)^{-2}. \]Practice 5: Radical and Chain Rule
Differentiate:
\[ f(x)=\sqrt{3x^2+1}. \]Practice 6: Product and Chain Rules
Differentiate:
\[ y=x^2(x+1)^5. \]Practice 7: Quotient and Chain Rules
Differentiate:
\[ y=\frac{(x^2+3)^2}{x-1}. \]Practice 8: Rewrite Before Differentiating
Differentiate efficiently:
\[ f(x)=\frac{1}{(2x+5)^3}. \]Practice 9: Nested Chain Rule
Differentiate:
\[ y=[(x^2+1)^3+2]^4. \]Practice 10: Identify the Error
A student differentiates
\[ y=(2x+1)^5 \]and writes
\[ y'=5(2x+1)^4. \]Identify the missing step and correct the derivative.
Practice Answers
Practice 1
Apply the product rule:
\[ y'=3x^2(x^2+5)+x^3(2x). \]Therefore,
\[ \boxed{y'=5x^4+15x^2}. \]Practice 2
\[ y' =\frac{(2x)(x+1)-(x^2-4)(1)}{(x+1)^2}. \]Simplify the numerator:
\[ 2x^2+2x-x^2+4=x^2+2x+4. \]Thus,
\[ \boxed{y'=\frac{x^2+2x+4}{(x+1)^2}}, \qquad x\ne-1. \]Practice 3
Differentiate the outer power and multiply by the inner derivative:
\[ y'=7(4x-3)^6(4). \]Therefore,
\[ \boxed{y'=28(4x-3)^6}. \]Practice 4
\[ f'(x) =-2(x^3+2x)^{-3}(3x^2+2). \]Equivalently,
\[ \boxed{ f'(x)=-\frac{2(3x^2+2)}{(x^3+2x)^3} } \]on intervals where the original expression is defined.
Practice 5
Rewrite as a power:
\[ f(x)=(3x^2+1)^{1/2}. \]Then
\[ f'(x)=\frac12(3x^2+1)^{-1/2}(6x). \]Therefore,
\[ \boxed{f'(x)=\frac{3x}{\sqrt{3x^2+1}}}. \]Practice 6
Apply the product rule, then the chain rule:
\[ y' =2x(x+1)^5+x^2\left[5(x+1)^4\right]. \]A useful factored form is
\[ \boxed{y'=x(x+1)^4(7x+2)}. \]Practice 7
The numerator derivative is
\[ \frac{d}{dx}[(x^2+3)^2] =2(x^2+3)(2x) =4x(x^2+3). \]Therefore,
\[ \boxed{ y'=\frac{4x(x^2+3)(x-1)-(x^2+3)^2}{(x-1)^2} }, \qquad x\ne1. \]Practice 8
Rewrite:
\[ f(x)=(2x+5)^{-3}. \]Then
\[ f'(x)=-3(2x+5)^{-4}(2). \]Thus,
\[ \boxed{f'(x)=-\frac{6}{(2x+5)^4}}. \]Practice 9
Differentiate each layer from the outside inward:
\[ y' =4[(x^2+1)^3+2]^3 \cdot3(x^2+1)^2 \cdot2x. \]Therefore,
\[ \boxed{ y'=24x[(x^2+1)^3+2]^3(x^2+1)^2 }. \]Practice 10
The student differentiated the outer fifth power but forgot to multiply by the derivative of the inner function \(2x+1\).
\[ \frac{d}{dx}(2x+1)=2. \]Therefore,
\[ \boxed{y'=10(2x+1)^4}. \]What to Remember
For a product,
\[ \boxed{(fg)'=f'g+fg'}. \]For a quotient,
\[ \boxed{ \left(\frac{f}{g}\right)'=\frac{f'g-fg'}{g^2} }, \qquad g\ne0. \]For a composition,
\[ \boxed{(f\circ g)'(x)=f'(g(x))g'(x)}. \]The most important working habit is to identify the function's structure before differentiating. Apply the rule for the outermost structure first, then move inward until every required derivative has been computed.
Where We Go Next
The rules in this lesson assume that the dependent quantity has been written explicitly enough for ordinary differentiation. Many important equations instead define variables indirectly, and some curves are described by giving both coordinates in terms of a third parameter.
The next lesson, Implicit and Parametric Differentiation, extends derivative methods to those forms and shows how the chain rule continues to operate even when the relationship is not written as a simple explicit function.