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The Chain Rule | Complete Guide (2026)

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The chain rule is one of the fundamental techniques of differentiation in calculus, used to compute the derivative of composite functions.
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The Chain Rule

Complete Guide to Calculus Chain Rule

Master the chain rule for differentiation and integration across Calculus 1, 2, 3, A-Level Mathematics, and advanced applications

The Chain Rule Formula

\[\frac{d}{dx}[f(g(x))] = f'(g(x)) \cdot g'(x)\]

🔗 Understanding the Chain Rule

🎯 What is the Chain Rule?

The chain rule is a fundamental theorem in calculus used to find the derivative of composite functions. When you have a function inside another function, like \(f(g(x))\), the chain rule tells you how to differentiate it step by step.

🧩 Composite Functions

A composite function is formed when one function is applied to the result of another function.

Example:

\[h(x) = (x^2 + 1)^3\]

Here, \(g(x) = x^2 + 1\) and \(f(u) = u^3\), so \(h(x) = f(g(x))\)

🎯 Why Do We Need It?

  • Complex Functions: Many real-world functions are composites
  • Efficiency: Faster than expanding and differentiating
  • Accuracy: Avoids algebraic errors in complex expressions
  • Generality: Works for any differentiable composite function
  • Foundation: Essential for advanced calculus topics

🔍 Identifying Composite Functions

Look for these patterns that indicate you need the chain rule:

  • Functions raised to powers: \((expression)^n\)
  • Functions inside trig functions: \(\sin(expression)\)
  • Functions inside exponentials: \(e^{expression}\)
  • Functions inside logarithms: \(\ln(expression)\)
  • Functions inside radicals: \(\sqrt{expression}\)

🎨 Intuitive Understanding

Think of the chain rule as the "rate of change of rate of change":

\[\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}\]

Rate of \(y\) with respect to \(x\) = (Rate of \(y\) w.r.t. \(u\)) × (Rate of \(u\) w.r.t. \(x\))

📐 Chain Rule Formula and Notation

🔑 The Chain Rule Formula

\[\boxed{\frac{d}{dx}[f(g(x))] = f'(g(x)) \cdot g'(x)}\]

In words: Derivative of outer function (evaluated at inner function) times derivative of inner function

📝 Alternative Notations

Leibniz Notation:

\[\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}\]

Prime Notation:

\[(f \circ g)'(x) = f'(g(x)) \cdot g'(x)\]

If \(y = f(u)\) and \(u = g(x)\):

\[\frac{dy}{dx} = \frac{df}{du} \cdot \frac{dg}{dx}\]

🎯 Step-by-Step Process

  1. Identify the outer function \(f\) and inner function \(g\)
  2. Find the derivative of the outer function: \(f'(u)\)
  3. Evaluate \(f'\) at the inner function: \(f'(g(x))\)
  4. Find the derivative of the inner function: \(g'(x)\)
  5. Multiply the results: \(f'(g(x)) \cdot g'(x)\)

🔄 Multiple Chain Rule

For functions with multiple compositions like \(f(g(h(x)))\):

\[\frac{d}{dx}[f(g(h(x)))] = f'(g(h(x))) \cdot g'(h(x)) \cdot h'(x)\]

Work from outside to inside, multiplying each derivative

⚡ Quick Reference

Common Chain Rule Patterns:

\[\frac{d}{dx}[u^n] = nu^{n-1} \cdot u'\] \[\frac{d}{dx}[e^u] = e^u \cdot u'\] \[\frac{d}{dx}[\ln u] = \frac{u'}{u}\] \[\frac{d}{dx}[\sin u] = \cos u \cdot u'\] \[\frac{d}{dx}[\cos u] = -\sin u \cdot u'\]

📝 Chain Rule Examples - Step by Step

🔢 Example 1: Power of a Function

Find the derivative: \(y = (3x^2 + 2x - 1)^5\)

Step 1: Identify outer and inner functions

  • Outer function: \(f(u) = u^5\), so \(f'(u) = 5u^4\)
  • Inner function: \(g(x) = 3x^2 + 2x - 1\), so \(g'(x) = 6x + 2\)

Step 2: Apply the chain rule

\[\frac{dy}{dx} = f'(g(x)) \cdot g'(x) = 5(3x^2 + 2x - 1)^4 \cdot (6x + 2)\]

✅ Final Answer:

\[\frac{dy}{dx} = 5(3x^2 + 2x - 1)^4(6x + 2)\]

📐 Example 2: Chain Rule with Trigonometric Functions

Find the derivative: \(y = \sin(x^3 + 2x)\)

Step 1: Identify functions

  • Outer function: \(f(u) = \sin u\), so \(f'(u) = \cos u\)
  • Inner function: \(g(x) = x^3 + 2x\), so \(g'(x) = 3x^2 + 2\)

Step 2: Apply the chain rule

\[\frac{dy}{dx} = \cos(x^3 + 2x) \cdot (3x^2 + 2)\]

✅ Final Answer:

\[\frac{dy}{dx} = (3x^2 + 2)\cos(x^3 + 2x)\]

📈 Example 3: Chain Rule with Exponential Functions

Find the derivative: \(y = e^{x^2 - 3x + 1}\)

Step 1: Identify functions

  • Outer function: \(f(u) = e^u\), so \(f'(u) = e^u\)
  • Inner function: \(g(x) = x^2 - 3x + 1\), so \(g'(x) = 2x - 3\)

Step 2: Apply the chain rule

\[\frac{dy}{dx} = e^{x^2 - 3x + 1} \cdot (2x - 3)\]

✅ Final Answer:

\[\frac{dy}{dx} = (2x - 3)e^{x^2 - 3x + 1}\]

🔗 Example 4: Multiple Chain Rule Application

Find the derivative: \(y = \cos^3(2x + 1) = [\cos(2x + 1)]^3\)

Step 1: Identify the three functions

  • Outermost: \(f(u) = u^3\), so \(f'(u) = 3u^2\)
  • Middle: \(g(v) = \cos v\), so \(g'(v) = -\sin v\)
  • Inner: \(h(x) = 2x + 1\), so \(h'(x) = 2\)

Step 2: Apply multiple chain rule

\[\frac{dy}{dx} = 3[\cos(2x + 1)]^2 \cdot (-\sin(2x + 1)) \cdot 2\]

✅ Final Answer:

\[\frac{dy}{dx} = -6\cos^2(2x + 1)\sin(2x + 1)\]

🚀 Advanced Chain Rule Applications

🔄 Implicit Differentiation

When \(y\) is defined implicitly, use the chain rule to find \(\frac{dy}{dx}\):

Example: \(x^2 + y^2 = 25\)

\[2x + 2y\frac{dy}{dx} = 0\] \[\frac{dy}{dx} = -\frac{x}{y}\]

Key: Treat \(y\) as a function of \(x\) and apply chain rule to \(y\) terms

∂ Chain Rule for Partial Derivatives

For multivariable functions \(z = f(x,y)\) where \(x = g(t)\) and \(y = h(t)\):

\[\frac{dz}{dt} = \frac{\partial f}{\partial x}\frac{dx}{dt} + \frac{\partial f}{\partial y}\frac{dy}{dt}\]

Essential for Calculus 3 and multivariable optimization

📊 Logarithmic Differentiation

For complex products, quotients, or variable exponents:

Example: \(y = x^x\)

\[\ln y = x \ln x\] \[\frac{1}{y}\frac{dy}{dx} = \ln x + 1\] \[\frac{dy}{dx} = x^x(\ln x + 1)\]

📈 Parametric Differentiation

For parametric equations \(x = f(t)\), \(y = g(t)\):

\[\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{g'(t)}{f'(t)}\]

Second derivative: \(\frac{d^2y}{dx^2} = \frac{d}{dx}\left(\frac{dy}{dx}\right) = \frac{d}{dt}\left(\frac{dy}{dx}\right) \cdot \frac{dt}{dx}\)

🎓 A-Level Chain Rule

Common A-Level applications include:

  • Composite trig functions: \(\sin(ax + b)\)
  • Exponential functions: \(e^{ax + b}\)
  • Logarithmic functions: \(\ln(ax + b)\)
  • Power functions: \((ax + b)^n\)
  • Product and quotient with chain rule

🤖 Chain Rule in Deep Learning

Backpropagation uses the chain rule to compute gradients:

\[\frac{\partial L}{\partial w} = \frac{\partial L}{\partial a} \cdot \frac{\partial a}{\partial z} \cdot \frac{\partial z}{\partial w}\]

Where \(L\) = loss, \(a\) = activation, \(z\) = weighted input, \(w\) = weight

∫ Chain Rule in Integration (Anti-Chain Rule)

🔄 Reverse Chain Rule (u-Substitution)

The "anti-chain rule" or u-substitution is used when integrating composite functions. If you can recognize a function and its derivative, you can integrate using substitution.

🎯 U-Substitution Method

Steps:

  1. Choose \(u\) (usually the inner function)
  2. Find \(du = u'dx\)
  3. Substitute to get integral in terms of \(u\)
  4. Integrate with respect to \(u\)
  5. Substitute back to get answer in terms of \(x\)

Pattern Recognition: Look for a function and its derivative (or a constant multiple)

📝 Integration Example 1

Evaluate: \(\int 2x(x^2 + 1)^3 dx\)

Let \(u = x^2 + 1\), then \(du = 2x dx\)

\[\int 2x(x^2 + 1)^3 dx = \int u^3 du = \frac{u^4}{4} + C\]
\[= \frac{(x^2 + 1)^4}{4} + C\]

📐 Integration Example 2

Evaluate: \(\int \cos(3x + 2) dx\)

Let \(u = 3x + 2\), then \(du = 3 dx\), so \(dx = \frac{du}{3}\)

\[\int \cos(3x + 2) dx = \int \cos u \cdot \frac{du}{3} = \frac{1}{3}\sin u + C\]
\[= \frac{1}{3}\sin(3x + 2) + C\]

⚡ Common Patterns

Recognizable Forms:

\[\int f(u) \cdot u' dx = \int f(u) du\] \[\int (ax + b)^n dx = \frac{(ax + b)^{n+1}}{a(n+1)} + C\] \[\int \frac{u'}{u} dx = \ln|u| + C\] \[\int u' e^u dx = e^u + C\]

⚠️ Common Mistakes & Pro Tips

❌ Common Mistakes

  • Forgetting the chain rule: Only differentiating the outer function
  • Wrong order: Multiplying \(g'(x) \cdot f'(g(x))\) instead of \(f'(g(x)) \cdot g'(x)\)
  • Not simplifying: Leaving answers in unnecessarily complex form
  • Incorrect identification: Misidentifying inner and outer functions
  • Integration confusion: Using chain rule formula for integration

✅ Pro Tips

  • Practice pattern recognition: Learn to spot composite functions quickly
  • Work inside-out: Always identify the innermost function first
  • Check your work: Differentiate your antiderivative to verify
  • Use parentheses: Keep track of function composition clearly
  • Master basics first: Know derivatives of elementary functions cold

🧠 Memory Aids

  • "Outside-inside": Differentiate outside, then multiply by inside derivative
  • "Chain of derivatives": Each link multiplies the next
  • "Peel the onion": Work from outer layer to inner layer
  • "Rate times rate": Rate of change times rate of change
  • Practice daily: 15 minutes of chain rule problems builds automaticity

📝 Practice Problems

🟢 Basic Level

  1. \(\frac{d}{dx}[(2x + 3)^4]\)
  2. \(\frac{d}{dx}[\sin(5x)]\)
  3. \(\frac{d}{dx}[e^{3x-1}]\)
  4. \(\frac{d}{dx}[\ln(x^2 + 1)]\)
  5. \(\frac{d}{dx}[\sqrt{4x + 7}]\)

🟡 Intermediate Level

  1. \(\frac{d}{dx}[\cos^2(x)]\)
  2. \(\frac{d}{dx}[e^{x^2}\sin(x)]\)
  3. \(\frac{d}{dx}[\ln(\cos(x))]\)
  4. \(\frac{d}{dx}[(x^2 + 1)^{3/2}]\)
  5. \(\frac{d}{dx}[\tan^{-1}(2x)]\)

🔴 Advanced Level

  1. \(\frac{d}{dx}[x^{\sin(x)}]\)
  2. \(\frac{d}{dx}[\sin^{-1}(\sqrt{x})]\)
  3. \(\frac{d}{dx}[e^{x^2}\cos(e^x)]\)
  4. Find \(\frac{dy}{dx}\) if \(x^2 + \sin(xy) = y^3\)
  5. \(\int \frac{2x}{x^2 + 1} dx\)

💡 Study Tip: Work through these problems step-by-step, clearly identifying inner and outer functions. Practice until the pattern recognition becomes automatic!

🎓 Chain Rule Across Different Curricula

📚 Calculus 1 Chain Rule

  • Basic composite function differentiation
  • Power rule combined with chain rule
  • Trigonometric function compositions
  • Exponential and logarithmic chains
  • Introduction to implicit differentiation

🧮 Calculus 3 Chain Rule

  • Multivariable chain rule
  • Partial derivative chains
  • Gradient and directional derivatives
  • Parametric and vector-valued functions
  • Applications in optimization

🇬🇧 A-Level Maths Chain Rule

  • Function of a function rule
  • Standard derivatives with chain rule
  • Connected rates of change
  • Integration by substitution
  • Applications to kinematics

🌍 IB Mathematics Chain Rule

  • Composite function differentiation
  • Applications to optimization problems
  • Implicit differentiation techniques
  • Integration using substitution
  • Real-world modeling applications

🇺🇸 AP Calculus Chain Rule

  • Fundamental theorem applications
  • Related rates problems
  • Optimization with constraints
  • FRQ problem-solving strategies
  • Calculator and non-calculator sections

📖 GCSE/IGCSE Preparation

  • Foundation differentiation skills
  • Basic composite function understanding
  • Preparation for advanced mathematics
  • Introduction to rate of change concepts
  • Building algebraic manipulation skills

👨‍🎓 About the Author

AK

Adam Kumar

Co-Founder @ RevisionTown

Adam is a mathematics education expert specializing in Calculus, Chain Rule, and Differentiation across various international curricula including IB Mathematics, AP Calculus, A-Level Mathematics, GCSE, IGCSE, and more. With years of experience helping students master the chain rule and its applications, he creates comprehensive, step-by-step study materials that make complex calculus concepts accessible and understandable.

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