Product Rule and Chain Rule for Derivatives

Product Rule and Chain Rule for Derivatives

When a function is built as a product of two factors, or as a composition of one function inside another, you cannot simply differentiate each part separately. On this page we introduce the two key tools: the product rule and the chain rule. We will develop the intuition behind each rule, learn when to apply which one, and practice computing derivatives and evaluating them at a point.

Background and Basic Definitions

When two functions are multiplied or composed, the derivative does not split apart in a simple way. That is why we have two central rules.

Product Rule — if \(h(x)=f(x)\cdot g(x)\) then:

\[ h'(x) = f'(x)\,g(x) + f(x)\,g'(x) \]

The intuition: each factor changes independently, so the total rate of change is the sum of two contributions — once we differentiate the first factor and keep the second, and once we keep the first and differentiate the second. A common mistake is to think \((f\cdot g)'=f'\cdot g'\) — that is incorrect.

Chain Rule — if \(h(x)=f\big(g(x)\big)\) (outer function \(f\) applied to inner function \(g\)) then:

\[ h'(x) = f'\big(g(x)\big)\cdot g'(x) \]

The intuition: differentiate the outer function as if the inner function were a single variable, then multiply by the derivative of the inner function. That multiplication corrects for the rate of change of the inner expression.

A common application of the chain rule is a power of an expression:

\[ \big[g(x)^n\big]' = n\,g(x)^{n-1}\cdot g'(x) \]

Recall the basic derivatives we will use: \((x^n)' = n x^{n-1}\), \((c)' = 0\) and \((c\cdot x)' = c\).

Basic Derivatives Table:

FunctionDerivativeFunctionDerivative
\(x^t\)\(t\cdot x^{t-1}\)\(\frac{1}{x}\)\(-\frac{1}{x^2}\)
\(\sqrt{x}\)\(\frac{1}{2\sqrt{x}}\)\(a^x\)\(a^x\cdot \ln a\)
\(\sin x\)\(\cos x\)\(\cos x\)\(-\sin x\)
\(\tan x\)\(\frac{1}{\cos^2 x}\)\(\log_a x\)\(\frac{1}{x\cdot \ln a}\)

where \(t\) is any real number. Differentiation rules: product \([f\cdot g]'=f'g+fg'\); quotient \(\left[\frac{f}{g}\right]'=\frac{f'g-fg'}{[g]^2}\); chain rule \([f(u(x))]'=f'(u)\cdot u'(x)\).

Solution Steps

  1. Step 1 — Identify the structure of the function: is it a product of two factors (product rule) or a composition of one function inside another (chain rule)?
  2. Step 2 (product) — Label \(f(x)\) and \(g(x)\), differentiate each separately to get \(f'(x)\) and \(g'(x)\).
  3. Step 3 (product) — Substitute into the formula \(h'=f'g+fg'\), expand the brackets and collect like terms.
  4. Step 4 (chain) — Identify the inner function \(g(x)\) and the outer function; differentiate the outer function evaluated at the inner function, then multiply by \(g'(x)\).
  5. Step 5 — If the value at a point is requested, substitute the \(x\)-value only at the end, after you have the derivative expression.
  6. Step 6 — Sanity check: the degree of the derivative is one less than the degree of the original function.

Worked Examples

Example 1: Product Rule — Two Linear Factors

Problem: Given the function \(h(x)=(2x+1)(x-3)\). Find \(h'(x)\).

Solution:

  1. Let \(f(x)=2x+1\) and \(g(x)=x-3\).
  2. Differentiate each factor: \(f'(x)=2\) and \(g'(x)=1\).
  3. Apply the product rule: \(h'(x)=f'g+fg'=2\,(x-3)+(2x+1)\cdot 1\).
  4. Expand: \(h'(x)=2x-6+2x+1\).
  5. Collect like terms: \(h'(x)=4x-5\).

Answer: \(h'(x)=4x-5\).

Example 2: Product Rule — Power Times Polynomial

Problem: Given the function \(h(x)=x^2(2x-5)\). Find \(h'(x)\).

Solution:

  1. Let \(f(x)=x^2\) and \(g(x)=2x-5\).
  2. Differentiate: \(f'(x)=2x\) and \(g'(x)=2\).
  3. Substitute: \(h'(x)=2x\,(2x-5)+x^2\cdot 2\).
  4. Expand: \(h'(x)=4x^2-10x+2x^2\).
  5. Collect: \(h'(x)=6x^2-10x\).

Answer: \(h'(x)=6x^2-10x\).

Example 3: Chain Rule — Power of a Linear Expression

Problem: Given the function \(h(x)=(3x+2)^4\). Find \(h'(x)\).

Solution:

  1. The inner function is \(g(x)=3x+2\) and the outer function is the fourth power.
  2. Differentiate the outer function: \(4(3x+2)^3\).
  3. Multiply by the derivative of the inner function \(g'(x)=3\).
  4. Result: \(h'(x)=4(3x+2)^3\cdot 3\).
  5. Simplify: \(h'(x)=12(3x+2)^3\).

Answer: \(h'(x)=12(3x+2)^3\).

Example 4: Chain Rule — Power of a Polynomial

Problem: Given the function \(h(x)=(x^2+1)^5\). Find \(h'(x)\).

Solution:

  1. The inner function is \(g(x)=x^2+1\) and the outer function is the fifth power.
  2. Differentiate the outer function: \(5(x^2+1)^4\).
  3. Differentiate the inner function: \(g'(x)=2x\).
  4. Multiply: \(h'(x)=5(x^2+1)^4\cdot 2x\).
  5. Simplify: \(h'(x)=10x(x^2+1)^4\).

Answer: \(h'(x)=10x(x^2+1)^4\).

Example 5: Product Rule — Evaluating the Derivative at a Point

Problem: Given the function \(h(x)=(x^2-2x)(x+3)\). Find \(h'(2)\).

Solution:

  1. Let \(f(x)=x^2-2x\) and \(g(x)=x+3\).
  2. Differentiate: \(f'(x)=2x-2\) and \(g'(x)=1\).
  3. Apply the product rule: \(h'(x)=(2x-2)(x+3)+(x^2-2x)\cdot 1\).
  4. Now substitute \(x=2\): \(h'(2)=(2\cdot 2-2)(2+3)+(2^2-2\cdot 2)\).
  5. Compute: \(h'(2)=(2)(5)+(4-4)=10+0=10\).

Answer: \(h'(2)=10\).

Common Mistakes

✗ Common mistake: Differentiating a product term by term and getting \((f\cdot g)'=f'\cdot g'\).

✓ The correct way: The derivative of a product is a sum: \(h'=f'g+fg'\). You must differentiate each factor separately while keeping the other factor intact.

✗ Common mistake: Forgetting to multiply by the derivative of the inner function when using the chain rule.

✓ The correct way: After differentiating the outer function you must multiply by \(g'(x)\). For example \([(3x+2)^4]'=4(3x+2)^3\cdot 3\), not just \(4(3x+2)^3\).

✗ Common mistake: Substituting the value of the point at the very beginning, before differentiating.

✓ The correct way: First find the derivative expression \(h'(x)\), then substitute the \(x\)-value at the end. Substituting early destroys the dependence on the variable.

Practice Tips

  • Tip — Before starting, ask yourself: 'Is this a product or a composition?' That determines which rule to apply.
  • Tip — With the product rule, you can verify your answer by expanding the brackets first and differentiating directly; both methods must give the same result.
  • Tip — With the chain rule, think 'outside in': first the power or root, then multiply by the 'inner derivative'.
  • Tip — When evaluating a derivative at a point, keep the expression in factored form (without expanding everything) — sometimes a factor becomes zero and the substitution is much easier.

Summary and Key Formulas

The two central rules:

  • Product rule: \(\big[f(x)g(x)\big]' = f'(x)g(x)+f(x)g'(x)\)
  • Chain rule: \(\big[f(g(x))\big]' = f'(g(x))\cdot g'(x)\)
  • Power of an expression: \(\big[g(x)^n\big]' = n\,g(x)^{n-1}\cdot g'(x)\)

To evaluate at a point — first find \(h'(x)\), then substitute the \(x\)-value.