🎯⭐ INTERACTIVE LESSON

Chain Rule

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Chain Rule - Complete Interactive Lesson

Part 1: Chain Rule Basics

🔗 The Chain Rule

Part 1 of 7 — Chain Rule Basics

Why Do We Need the Chain Rule?

So far, you can differentiate functions like x3x^3, sinx\sin x, or exe^x. But what about composite functions — functions inside other functions?

Consider f(x)=(3x+1)5f(x) = (3x + 1)^5. You could expand this, but that is painful. What about sin(x2)\sin(x^2) or e3xe^{3x}? There is no shortcut without the Chain Rule.

The Chain Rule Formula

If y=f(g(x))y = f(g(x)), then:

dydx=f(g(x))g(x)\frac{dy}{dx} = f'(g(x)) \cdot g'(x)

In words: differentiate the outer function (leaving the inner function untouched), then multiply by the derivative of the inner function.

Leibniz Notation

If y=f(u)y = f(u) where u=g(x)u = g(x), then:

dydx=dydududx\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}


Worked Example 1

Find ddx(3x+1)5\frac{d}{dx}(3x+1)^5

StepWork
Identify layersOuter: u5u^5, Inner: u=3x+1u = 3x+1
Differentiate outer5u4=5(3x+1)45u^4 = 5(3x+1)^4
Differentiate innerddx(3x+1)=3\frac{d}{dx}(3x+1) = 3
Multiply5(3x+1)43=15(3x+1)45(3x+1)^4 \cdot 3 = 15(3x+1)^4

Worked Example 2

Find ddxsin(x2)\frac{d}{dx}\sin(x^2)

StepWork
Identify layersOuter: sin(u)\sin(u), Inner: u=x2u = x^2
Differentiate outercos(u)=cos(x2)\cos(u) = \cos(x^2)
Differentiate innerddx(x2)=2x\frac{d}{dx}(x^2) = 2x
Multiplycos(x2)2x=2xcos(x2)\cos(x^2) \cdot 2x = 2x\cos(x^2)

AP Tip: The Chain Rule appears in nearly every derivative problem on the AP exam. Master it now.

Check Your Understanding 🎯

Compute the derivative using the Chain Rule.

Worked Example 3

Find ddxx2+1\frac{d}{dx}\sqrt{x^2 + 1}

Rewrite: x2+1=(x2+1)1/2\sqrt{x^2+1} = (x^2+1)^{1/2}

StepWork
Outer derivative12(x2+1)1/2\frac{1}{2}(x^2+1)^{-1/2}
Inner derivative2x2x
Chain Rule12(x2+1)1/22x=xx2+1\frac{1}{2}(x^2+1)^{-1/2} \cdot 2x = \frac{x}{\sqrt{x^2+1}}

Worked Example 4

Find ddxex2\frac{d}{dx}e^{-x^2}

StepWork
Outer: eue^uex2e^{-x^2}
Inner: u=x2u = -x^22x-2x
Chain Ruleex2(2x)=2xex2e^{-x^2} \cdot (-2x) = -2xe^{-x^2}

Apply the Chain Rule 🎯

Identify the Outer Function 🔍

For each composite function, select the correct outer function.

Key Takeaways — Part 1

FunctionDerivative
(stuff)n(\text{stuff})^nn(stuff)n1(stuff)n(\text{stuff})^{n-1} \cdot (\text{stuff})'
sin(stuff)\sin(\text{stuff})cos(stuff)(stuff)\cos(\text{stuff}) \cdot (\text{stuff})'
cos(stuff)\cos(\text{stuff})sin(stuff)(stuff)-\sin(\text{stuff}) \cdot (\text{stuff})'
estuffe^{\text{stuff}}estuff(stuff)e^{\text{stuff}} \cdot (\text{stuff})'
ln(stuff)\ln(\text{stuff})1stuff(stuff)\frac{1}{\text{stuff}} \cdot (\text{stuff})'

Part 2: Nested Functions

🔗 Nested Functions & Double Chain Rule

Part 2 of 7 — Nested Functions

When the Chain Rule Applies Twice

Some functions have three or more layers. For example:

f(x)=sin2(3x)=[sin(3x)]2f(x) = \sin^2(3x) = [\sin(3x)]^2

Here we have three layers:

  1. Outermost: u2u^2
  2. Middle: sin(v)\sin(v)
  3. Innermost: v=3xv = 3x

The Chain Rule applies at each layer, multiplying all the derivatives together.

Worked Example 1

Find ddx[sin(3x)]2\frac{d}{dx}[\sin(3x)]^2

ddx[sin(3x)]2=2sin(3x)cos(3x)3=6sin(3x)cos(3x)\frac{d}{dx}[\sin(3x)]^2 = 2\sin(3x) \cdot \cos(3x) \cdot 3 = 6\sin(3x)\cos(3x)

LayerFunctionDerivative
Outeru2u^22u=2sin(3x)2u = 2\sin(3x)
Middlesin(v)\sin(v)cos(v)=cos(3x)\cos(v) = \cos(3x)
Inner3x3x33
Result2sin(3x)cos(3x)3=6sin(3x)cos(3x)2\sin(3x) \cdot \cos(3x) \cdot 3 = 6\sin(3x)\cos(3x)

Worked Example 2

Find ddxecos(2x)\frac{d}{dx}e^{\cos(2x)}

LayerFunctionDerivative
Outereue^uecos(2x)e^{\cos(2x)}
Middlecos(v)\cos(v)sin(2x)-\sin(2x)
Inner2x2x22
Resultecos(2x)(sin(2x))2=2sin(2x)ecos(2x)e^{\cos(2x)} \cdot (-\sin(2x)) \cdot 2 = -2\sin(2x)e^{\cos(2x)}

Worked Example 3

Find ddxln(x)\frac{d}{dx}\sqrt{\ln(x)}

Rewrite as (lnx)1/2(\ln x)^{1/2}.

LayerFunctionDerivative
Outeru1/2u^{1/2}12(lnx)1/2\frac{1}{2}(\ln x)^{-1/2}
Innerlnx\ln x1x\frac{1}{x}
Result12lnx1x=12xlnx\frac{1}{2\sqrt{\ln x}} \cdot \frac{1}{x} = \frac{1}{2x\sqrt{\ln x}}

Differentiate these nested functions 🎯

Triple Chain Rule

Sometimes you have four layers. The process is the same — just keep peeling.

Worked Example 4

Find ddxsin3(2x+1)\frac{d}{dx}\sin^3(2x+1)

This means [sin(2x+1)]3[\sin(2x+1)]^3:

LayerDerivative
u3u^33[sin(2x+1)]23[\sin(2x+1)]^2
sin(v)\sin(v)cos(2x+1)\cos(2x+1)
2x+12x+122
Result3sin2(2x+1)cos(2x+1)2=6sin2(2x+1)cos(2x+1)3\sin^2(2x+1) \cdot \cos(2x+1) \cdot 2 = 6\sin^2(2x+1)\cos(2x+1)

Common Mistake Alert ⚠️

Students often forget the innermost derivative. For sin2(3x)\sin^2(3x), many write 2sin(3x)cos(3x)2\sin(3x)\cos(3x) and forget the factor of 33.

Always ask: "Is there another layer inside that I haven't differentiated yet?"

Multi-Layer Problems 🎯

How many Chain Rule applications? 🔍

For each function, select how many times you must apply the Chain Rule.

Key Takeaways — Part 2

  1. Nested functions require the Chain Rule applied multiple times
  2. Multiply ALL layer derivatives together — do not stop early
  3. Common mistake: forgetting the innermost derivative
  4. Strategy: Write out each layer, differentiate each, then multiply

Next up: Implicit differentiation — using the Chain Rule when yy is a function of xx.

Part 3: Implicit Differentiation

🔗 Implicit Differentiation

Part 3 of 7 — Implicit Differentiation

What Is Implicit Differentiation?

Sometimes a relationship between xx and yy is not solved for yy. For example:

x2+y2=25x^2 + y^2 = 25

This is a circle. We cannot easily write yy as a single function of xx. But we can still find dydx\frac{dy}{dx} using the Chain Rule.

The Key Idea

When you differentiate a term containing yy with respect to xx, treat yy as a function of xx and apply the Chain Rule:

ddx[y2]=2ydydx\frac{d}{dx}[y^2] = 2y \cdot \frac{dy}{dx}

The dydx\frac{dy}{dx} appears because yy is implicitly a function of xx.

Worked Example 1

Find dydx\frac{dy}{dx} for x2+y2=25x^2 + y^2 = 25

StepWork
Differentiate both sides2x+2ydydx=02x + 2y\frac{dy}{dx} = 0
Isolate dydx\frac{dy}{dx}2ydydx=2x2y\frac{dy}{dx} = -2x
Solvedydx=xy\frac{dy}{dx} = -\frac{x}{y}

Worked Example 2

Find dydx\frac{dy}{dx} for x3+y3=6xyx^3 + y^3 = 6xy

StepWork
Differentiate3x2+3y2dydx=6y+6xdydx3x^2 + 3y^2\frac{dy}{dx} = 6y + 6x\frac{dy}{dx}
Group dy/dxdy/dx terms3y2dydx6xdydx=6y3x23y^2\frac{dy}{dx} - 6x\frac{dy}{dx} = 6y - 3x^2
Factordydx(3y26x)=6y3x2\frac{dy}{dx}(3y^2 - 6x) = 6y - 3x^2
Solvedydx=6y3x23y26x=2yx2y22x\frac{dy}{dx} = \frac{6y - 3x^2}{3y^2 - 6x} = \frac{2y - x^2}{y^2 - 2x}

AP Tip: Implicit differentiation appears frequently on the AP exam, especially when finding slopes of tangent lines to curves defined implicitly.

Practice Implicit Differentiation 🎯

Implicit Differentiation with Trig Functions

Worked Example 3

Find dydx\frac{dy}{dx} for sin(y)=x\sin(y) = x

StepWork
Differentiatecos(y)dydx=1\cos(y) \cdot \frac{dy}{dx} = 1
Solvedydx=1cos(y)\frac{dy}{dx} = \frac{1}{\cos(y)}

This also equals sec(y)\sec(y), which makes sense since y=arcsin(x)y = \arcsin(x) and ddxarcsin(x)=11x2=sec(arcsin(x))\frac{d}{dx}\arcsin(x) = \frac{1}{\sqrt{1-x^2}} = \sec(\arcsin(x)).

Worked Example 4: Finding a Tangent Line

Find the slope of the tangent line to x2+xy+y2=7x^2 + xy + y^2 = 7 at (1,2)(1, 2).

StepWork
Differentiate2x+y+xdydx+2ydydx=02x + y + x\frac{dy}{dx} + 2y\frac{dy}{dx} = 0
Collect dy/dxdy/dxdydx(x+2y)=2xy\frac{dy}{dx}(x + 2y) = -2x - y
Solvedydx=2xyx+2y\frac{dy}{dx} = \frac{-2x - y}{x + 2y}
Evaluate at (1,2)(1,2)dydx=2(1)21+2(2)=45\frac{dy}{dx} = \frac{-2(1) - 2}{1 + 2(2)} = \frac{-4}{5}

The tangent line has slope 45-\frac{4}{5} at (1,2)(1, 2).

Implicit Differentiation Applications 🎯

Which differentiation rule is needed? 🔍

For each term, select the rule needed when differentiating with respect to xx.

Key Takeaways — Part 3

  1. Implicit differentiation uses the Chain Rule: every time you differentiate yy, multiply by dydx\frac{dy}{dx}
  2. Steps: differentiate both sides → collect dydx\frac{dy}{dx} terms → factor → solve
  3. Product rule is often needed when xx and yy appear together (like xyxy)
  4. Tangent lines: plug in the point after finding the general dydx\frac{dy}{dx} formula

Next up: Related rates — using implicit differentiation with respect to time.

Part 4: Related Rates Intro

🔗 Related Rates

Part 4 of 7 — Related Rates Intro

What Are Related Rates?

In related rates problems, two or more quantities are changing with respect to time (tt), and they are connected by an equation. We use implicit differentiation (with respect to tt) to find how fast one quantity changes given information about the other.

The Strategy

  1. Draw a picture and label changing quantities with variables
  2. Write an equation relating the variables
  3. Differentiate both sides with respect to tt (implicit differentiation)
  4. Substitute known values and solve for the unknown rate

Worked Example 1: Expanding Circle

A stone is dropped in a pond. The circular ripple expands so that its radius increases at 22 ft/s. How fast is the area increasing when the radius is 55 ft?

StepWork
Knowndrdt=2\frac{dr}{dt} = 2 ft/s, r=5r = 5 ft
FinddAdt\frac{dA}{dt}
EquationA=πr2A = \pi r^2
DifferentiatedAdt=2πrdrdt\frac{dA}{dt} = 2\pi r \cdot \frac{dr}{dt}
SubstitutedAdt=2π(5)(2)=20π62.8\frac{dA}{dt} = 2\pi(5)(2) = 20\pi \approx 62.8 ft2^2/s

Worked Example 2: Ladder Problem

A 13-ft ladder leans against a wall. The bottom slides away from the wall at 2 ft/s. How fast is the top sliding down when the bottom is 5 ft from the wall?

StepWork
Setupx2+y2=169x^2 + y^2 = 169 (Pythagorean theorem)
Knowndxdt=2\frac{dx}{dt} = 2 ft/s, x=5x = 5
Find yyy=16925=12y = \sqrt{169 - 25} = 12
Differentiate2xdxdt+2ydydt=02x\frac{dx}{dt} + 2y\frac{dy}{dt} = 0
Substitute2(5)(2)+2(12)dydt=02(5)(2) + 2(12)\frac{dy}{dt} = 0
Solvedydt=2024=56\frac{dy}{dt} = -\frac{20}{24} = -\frac{5}{6} ft/s

The negative sign means the top is sliding down at 56\frac{5}{6} ft/s.

Solve These Related Rates Problems 🎯

Worked Example 3: Conical Tank

Water drains from a conical tank (vertex down) at 2 ft3^3/min. The cone has radius 3 ft and height 6 ft at the top. How fast is the water level dropping when the depth is 4 ft?

Since the cone is similar: rh=36=12\frac{r}{h} = \frac{3}{6} = \frac{1}{2}, so r=h2r = \frac{h}{2}.

StepWork
Volume formulaV=13πr2h=13π(h2)2h=πh312V = \frac{1}{3}\pi r^2 h = \frac{1}{3}\pi \left(\frac{h}{2}\right)^2 h = \frac{\pi h^3}{12}
DifferentiatedVdt=π4h2dhdt\frac{dV}{dt} = \frac{\pi}{4} h^2 \frac{dh}{dt}
Substitute2=π4(16)dhdt-2 = \frac{\pi}{4}(16)\frac{dh}{dt}
Solvedhdt=24π=12π\frac{dh}{dt} = \frac{-2}{4\pi} = -\frac{1}{2\pi} ft/min

AP Tip: Related rates problems are a staple of AP Calculus free-response questions. Always clearly state what each variable represents and what rate you are finding.

More Related Rates 🎯

Key Takeaways — Part 4

  1. Related rates = implicit differentiation with respect to time
  2. Steps: draw → equation → differentiate → substitute → solve
  3. Key equations to know: Pythagorean theorem, area/volume formulas, similar triangles
  4. Watch signs: negative rates mean decreasing quantities

Next up: More Chain Rule applications including logarithmic differentiation.

Part 5: Applications

🔗 Advanced Chain Rule Applications

Part 5 of 7 — Applications

Logarithmic Differentiation

For functions like y=xxy = x^x or y=(sinx)cosxy = (\sin x)^{\cos x}, standard rules fail. Logarithmic differentiation solves this:

  1. Take ln\ln of both sides
  2. Use log properties to simplify
  3. Differentiate implicitly
  4. Solve for dydx\frac{dy}{dx}

Worked Example 1

Find ddxxx\frac{d}{dx}x^x for x>0x > 0

StepWork
Take lnlny=xlnx\ln y = x \ln x
Differentiate1ydydx=lnx+x1x=lnx+1\frac{1}{y}\frac{dy}{dx} = \ln x + x \cdot \frac{1}{x} = \ln x + 1
Solvedydx=y(lnx+1)=xx(lnx+1)\frac{dy}{dx} = y(\ln x + 1) = x^x(\ln x + 1)

Worked Example 2: Simplifying Products

Find ddxx2x+1(2x3)4\frac{d}{dx}\frac{x^2\sqrt{x+1}}{(2x-3)^4}

Without log differentiation, you would need the quotient rule, product rule, and chain rule combined. With logs:

StepWork
Take lnlny=2lnx+12ln(x+1)4ln(2x3)\ln y = 2\ln x + \frac{1}{2}\ln(x+1) - 4\ln(2x-3)
Differentiateyy=2x+12(x+1)82x3\frac{y'}{y} = \frac{2}{x} + \frac{1}{2(x+1)} - \frac{8}{2x-3}
Solvey=x2x+1(2x3)4(2x+12(x+1)82x3)y' = \frac{x^2\sqrt{x+1}}{(2x-3)^4}\left(\frac{2}{x} + \frac{1}{2(x+1)} - \frac{8}{2x-3}\right)

Logarithmic Differentiation 🎯

Chain Rule with Inverse Trig Functions

Recall the derivatives of inverse trig functions — they all involve the Chain Rule when the argument is not just xx:

FunctionDerivative
arcsin(u)\arcsin(u)u1u2\frac{u'}{\sqrt{1-u^2}}
arccos(u)\arccos(u)u1u2\frac{-u'}{\sqrt{1-u^2}}
arctan(u)\arctan(u)u1+u2\frac{u'}{1+u^2}

Worked Example 3

Find ddxarctan(3x)\frac{d}{dx}\arctan(3x)

ddxarctan(3x)=31+(3x)2=31+9x2\frac{d}{dx}\arctan(3x) = \frac{3}{1+(3x)^2} = \frac{3}{1+9x^2}

Worked Example 4

Find ddxarcsin(x2)\frac{d}{dx}\arcsin(x^2)

ddxarcsin(x2)=2x1(x2)2=2x1x4\frac{d}{dx}\arcsin(x^2) = \frac{2x}{\sqrt{1-(x^2)^2}} = \frac{2x}{\sqrt{1-x^4}}

Inverse Trig Derivatives 🎯

Key Takeaways — Part 5

  1. Logarithmic differentiation handles uvu^v where both base and exponent are functions of xx
  2. Steps: ln\ln both sides → simplify → differentiate implicitly → solve
  3. Inverse trig + Chain Rule: always multiply by the derivative of the inner function
  4. The Chain Rule is truly everywhere in calculus — almost no real problem avoids it

Next up: Problem-Solving Workshop with mixed Chain Rule problems.

Part 6: Problem-Solving Workshop

🔗 Problem-Solving Workshop

Part 6 of 7 — Mixed Chain Rule Problems

Strategy Recap

For any derivative problem:

  1. Identify if it is a composite function (Chain Rule needed?)
  2. Count layers — how many times must the Chain Rule be applied?
  3. Check for Product Rule or Quotient Rule requirements as well
  4. Apply rules carefully, and do not forget the innermost derivative

This workshop tests your ability to combine the Chain Rule with other rules in realistic AP-style problems.

AP-Style Problems — Set 1 🎯

AP-Style Problems — Set 2 🎯

Match the derivative 🔍

Select the correct derivative for each function.

Workshop Complete!

You have practiced combining the Chain Rule with:

  • Product Rule
  • Quotient Rule
  • Implicit differentiation
  • Logarithmic differentiation
  • Inverse trig functions

Next up: Review and comprehensive assessment.

Part 7: Review & Applications

🔗 Chain Rule Review

Part 7 of 7 — Review & Applications

Complete Chain Rule Summary

ScenarioFormula
Basic Chain Ruleddx[f(g(x))]=f(g(x))g(x)\frac{d}{dx}[f(g(x))] = f'(g(x)) \cdot g'(x)
Nested (double)Multiply ALL layer derivatives
ImplicitDifferentiate yy terms, attach dydx\frac{dy}{dx}
Related RatesDifferentiate with respect to tt
Log DifferentiationTake ln\ln, differentiate implicitly

AP Exam Frequency

The Chain Rule appears in:

  • ~80% of all derivative problems (it is rarely absent)
  • All implicit differentiation problems
  • All related rates problems
  • FTC Part 1 problems involving ddxag(x)f(t)dt\frac{d}{dx}\int_a^{g(x)} f(t)\,dt

Comprehensive Assessment 🎯

No hints — test your mastery.

Final Challenge 🎯

Chain Rule — Complete! ✅

You have mastered:

  • ✅ Basic Chain Rule with single composition
  • ✅ Nested functions requiring multiple applications
  • ✅ Implicit differentiation
  • ✅ Related rates
  • ✅ Logarithmic differentiation
  • ✅ Inverse trig with Chain Rule
  • ✅ Combining Chain Rule with Product and Quotient Rules

You are ready to tackle any Chain Rule problem on the AP exam!