Unit 3: Differentiation: Composite, Implicit, and Inverse Functions

The Chain Rule for Composite Functions

What “composite” means and why derivatives get trickier

A composite function is a function built by feeding one function into another. In function notation, that looks like f(g(x)), read “f of g(x).” Here, g(x) is the inside function (it produces the input), and f is the outside function (it acts on whatever comes out of g).

Derivatives become more interesting with composites because a small change in x first causes a change in g(x), and then that change in g(x) causes a change in f(g(x)). So the overall rate of change is a product of two rates: how fast f changes with respect to its input, and how fast g changes with respect to x.

The chain rule (what it is)

The chain rule is the differentiation rule for composite functions. If y=f(g(x)) and both functions are differentiable, then

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

A common way to make the roles clearer is to use an intermediate variable. Let u=g(x) and y=f(u). Then

\frac{dy}{dx}=\frac{dy}{du}\frac{du}{dx}

A useful memory phrase is “derivative of the outer, leave the inner, multiply by the derivative of the inner.” Another popular mnemonic is “douter, inner, dinner”: differentiate the outside first (often “drop the power” if it’s a power function), keep the inside expression as-is, then multiply by the inside derivative.

Why it matters (big picture)

The chain rule is what lets you differentiate almost anything you can write with standard functions, such as \sin(x^2), e^{\cos x}, or nested expressions like \sqrt{1+\sqrt{x}}. It also shows up in applied problems whenever one quantity depends on a second quantity, which depends on a third.

How to use it (a reliable step-by-step process)

When you see a function that is “something done to something,” use this process.

1. Identify the outside operation (the last thing being done).
2. Differentiate the outside, treating the inside as a single unit.
3. Multiply by the derivative of the inside.

A helpful mindset is that every time you differentiate an expression with an “inside,” you should expect an extra factor from differentiating that inside.

Common chain rule patterns

If y=(g(x))^n, then

y'=n(g(x))^{n-1}g'(x)

If y=\sin(g(x)), then

y'=\cos(g(x))g'(x)

If y=e^{g(x)}, then

y'=e^{g(x)}g'(x)

If y=\ln(g(x)), then

y'=\frac{g'(x)}{g(x)}

Worked example 1: a single-layer composite

Differentiate

y=\sin(3x^2-5x)

The outside function is \sin(\cdot) and the inside is g(x)=3x^2-5x. Using the chain rule,

y'=\cos(3x^2-5x)(6x-5)

A common mistake is to stop at \cos(3x^2-5x). The chain rule requires multiplying by the inside derivative.

Worked example 2: multiple layers (chain rule more than once)

Differentiate

y=\sqrt{1+(2x-1)^4}

Rewrite as

y=(1+(2x-1)^4)^{1/2}

Differentiate the outer layer, then the inside (which itself requires chain rule):

y'=\frac{1}{2}(1+(2x-1)^4)^{-1/2}\cdot 8(2x-1)^3

So

y'=4(2x-1)^3(1+(2x-1)^4)^{-1/2}

Equivalently,

y'=\frac{4(2x-1)^3}{\sqrt{1+(2x-1)^4}}

Chain rule with tables and function values

On AP-style problems, you may be given values like g(2)=5, g'(2)=-1, f'(5)=3 and asked for \frac{d}{dx}f(g(x)) at x=2. Use

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

Then substitute in the correct order. For example,

\frac{d}{dx}f(g(x))\bigg|_{x=2}=f'(g(2))g'(2)=f'(5)(-1)=-3

A common error is to use f'(2) instead of f'(g(2)).

Exam Focus

• Typical question patterns:
  • Differentiate expressions with nested functions (often involving trig, radicals, exponentials, or logs).
  • Evaluate \frac{d}{dx}f(g(x)) at a point using a table of values for f', g, and g'.
  • Combine chain rule with product or quotient rules in one derivative.
• Common mistakes:
  • Forgetting to multiply by the inside derivative.
  • Misidentifying the inside when there are parentheses, exponents, or roots.
  • In table problems, plugging the wrong input into f' by using x instead of g(x).

Implicit Differentiation

What “implicit” means

An equation is implicit when y is not isolated as a function of x. For example,

x^2+y^2=25

defines a relationship between x and y without explicitly writing y=\dots. Implicit differentiation means differentiating both sides with respect to x while treating y as a function of x. This is a favorite AP exam skill because it forces you to combine rules carefully.

Why it matters

Implicit differentiation lets you find slopes of curves that fail the vertical line test (like circles), compute tangent lines without solving explicitly for y, and differentiate relationships where solving for y first is difficult.

The core idea: derivatives of y require \frac{dy}{dx}

Because y=y(x), anything involving y must be chain-ruled when differentiating with respect to x. A helpful way to say this is: if the variable inside a function doesn’t match dx, you must follow it with the derivative of that variable with respect to x.

Examples:

\frac{d}{dx}[y]=\frac{dy}{dx}

\frac{d}{dx}[y^2]=2y\frac{dy}{dx}

\frac{d}{dx}[\sin(y)]=\cos(y)\frac{dy}{dx}

How implicit differentiation works (step-by-step)

1. Differentiate both sides with respect to x.
2. Whenever you differentiate a term containing y, multiply by \frac{dy}{dx}.
3. Collect all terms with \frac{dy}{dx} on one side.
4. Factor out \frac{dy}{dx} and solve.

Worked example 1: circle (including slope and tangent line)

Find \frac{dy}{dx} if

x^2+y^2=25

Differentiate:

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

Solve:

\frac{dy}{dx}=-\frac{x}{y}

At the point (3,4),

\frac{dy}{dx}\bigg|_{(3,4)}=-\frac{3}{4}

So the tangent line (point-slope form) is

y-4=-\frac{3}{4}(x-3)

Worked example 2: product and trig in an implicit equation

Find \frac{dy}{dx} if

xy+\sin(y)=x^2

Differentiate. For xy, use the product rule:

\frac{d}{dx}[xy]=x\frac{dy}{dx}+y

For \sin(y), use chain rule:

\frac{d}{dx}[\sin(y)]=\cos(y)\frac{dy}{dx}

So

x\frac{dy}{dx}+y+\cos(y)\frac{dy}{dx}=2x

Group and solve:

\frac{dy}{dx}=\frac{2x-y}{x+\cos(y)}

A frequent mistake is differentiating xy as only y or only x\frac{dy}{dx}.

Using \frac{dx}{dy} and reciprocals (when helpful)

Sometimes a problem or your algebra naturally produces \frac{dx}{dy} instead of \frac{dy}{dx}. The derivatives are reciprocals (when both are defined):

\frac{dx}{dy}=\frac{1}{\frac{dy}{dx}}

In some setups, solving for \frac{dx}{dy} first can make factoring and isolating the derivative cleaner, and then you can take the reciprocal to get \frac{dy}{dx}.

Second derivative with implicit differentiation

To find \frac{d^2y}{dx^2}, first find \frac{dy}{dx}, then differentiate that expression with respect to x again, remembering that y still depends on x.

From the circle,

\frac{dy}{dx}=-\frac{x}{y}

Differentiate:

\frac{d^2y}{dx^2}=\frac{d}{dx}\left[-\frac{x}{y}\right]

Using the quotient rule with u=-x and v=y gives

\frac{d^2y}{dx^2}=\frac{-y+x\frac{dy}{dx}}{y^2}

Substitute \frac{dy}{dx}=-\frac{x}{y}:

\frac{d^2y}{dx^2}=-\frac{y^2+x^2}{y^3}

Using the original equation x^2+y^2=25,

\frac{d^2y}{dx^2}=-\frac{25}{y^3}

A strong practical habit is to simplify your first-derivative expression before taking a second derivative.

Exam Focus

• Typical question patterns:
  • Differentiate an implicit equation and solve for \frac{dy}{dx}.
  • Find the slope of the tangent line at a given point on an implicit curve.
  • Compute \frac{d^2y}{dx^2} for an implicit relation.
• Common mistakes:
  • Forgetting \frac{dy}{dx} when differentiating terms involving y.
  • Incorrectly differentiating products like xy (missing product rule).
  • Plugging in a point too early (before solving for \frac{dy}{dx}), which can create algebra traps.

Derivatives of Inverse Functions (General)

What an inverse function is

An inverse function undoes another function. If f and f^{-1} are inverses, then

f(f^{-1}(x))=x

and

f^{-1}(f(x))=x

This only works on intervals where f is one-to-one.

Why inverses affect derivatives

Inverse functions reflect a graph across the line y=x, which swaps input and output. That reflection swaps the roles of “rise” and “run,” so slopes become reciprocals (where they exist). Also, points swap coordinates: if (a,b) is on f, then (b,a) is on f^{-1}.

The inverse derivative formula

If f is differentiable and one-to-one on an interval, and f'(a)\ne 0, then

\left(f^{-1}\right)'(f(a))=\frac{1}{f'(a)}

Equivalently,

\left(f^{-1}\right)'(b)=\frac{1}{f'(f^{-1}(b))}

A practical way to remember the point correspondence is: to find the slope of the inverse at input b, first find the matching original input a=f^{-1}(b), then take the reciprocal of f'(a).

Where the formula comes from (conceptual derivation)

Start with

f(f^{-1}(x))=x

Differentiate:

f'(f^{-1}(x))\left(f^{-1}\right)'(x)=1

So

\left(f^{-1}\right)'(x)=\frac{1}{f'(f^{-1}(x))}

This highlights two key details: you evaluate f' at f^{-1}(x) (not at x), and the inverse derivative exists only where that denominator is nonzero.

Worked example 1: using a point and a derivative

Suppose f(2)=7 and f'(2)=5. Because f^{-1}(7)=2,

\left(f^{-1}\right)'(7)=\frac{1}{5}

Worked example 2: computing \left(f^{-1}\right)'(b) from an explicit function

Let

f(x)=x^3+1

Find \left(f^{-1}\right)'(9). First solve x^3+1=9, so x=2 and f^{-1}(9)=2. Then f'(x)=3x^2, so f'(2)=12. Therefore,

\left(f^{-1}\right)'(9)=\frac{1}{12}

A quick point-swap reminder (common “at a point” phrasing)

If you are told you want the derivative of the inverse at a point like (1,2) on the inverse, that corresponds to the point (2,1) on the original function. In that situation, you would find f'(2) and then take the reciprocal to get the inverse slope at x=1.

A quick notation reference (to avoid confusion)

Be careful: f^{-1}(x) does not mean \frac{1}{f(x)}.

Exam Focus

• Typical question patterns:
  • Given f(a)=b and f'(a), find \left(f^{-1}\right)'(b).
  • Use f(f^{-1}(x))=x and differentiate (sometimes embedded in a larger problem).
  • Determine where \left(f^{-1}\right)'(x) exists based on where f'(x)\ne 0.
• Common mistakes:
  • Confusing f^{-1}(x) with \frac{1}{f(x)}.
  • Plugging b into f' instead of finding f^{-1}(b) first.
  • Ignoring the condition f'(a)\ne 0.
• AP pacing note:
  • These questions are often just 1 to 2 items on an exam, so aim for accuracy and a clean procedure rather than overcomplicating.

Inverse Trigonometric Functions and Their Derivatives

What inverse trig functions actually mean

Inverse trig functions answer questions like “what angle has this sine value?” For example, \arcsin(x) is the angle (in a chosen range) whose sine is x. Because trig functions are periodic, they are not one-to-one on their full domains, so we restrict them to principal ranges.

Common principal ranges:

• y=\arcsin(x) returns angles in

\left[-\frac{\pi}{2},\frac{\pi}{2}\right]

• y=\arccos(x) returns angles in

[0,\pi]

• y=\arctan(x) returns angles in

\left(-\frac{\pi}{2},\frac{\pi}{2}\right)

Why their derivatives matter

Inverse trig derivatives show up in derivatives like \arctan(3x) and \arcsin(x^2), in later integration topics, and in implicit differentiation outcomes that simplify to inverse trig forms. Many students choose to memorize these, but you can also derive them using implicit differentiation and trig identities.

Derivatives of the big three (core AP set)

\frac{d}{dx}[\arcsin(x)]=\frac{1}{\sqrt{1-x^2}}

\frac{d}{dx}[\arccos(x)]=-\frac{1}{\sqrt{1-x^2}}

\frac{d}{dx}[\arctan(x)]=\frac{1}{1+x^2}

With input u=g(x), apply the chain rule:

\frac{d}{dx}[\arcsin(g(x))]=\frac{g'(x)}{\sqrt{1-(g(x))^2}}

\frac{d}{dx}[\arctan(g(x))]=\frac{g'(x)}{1+(g(x))^2}

Where these come from (one example via implicit differentiation)

Let

y=\arcsin(x)

Then

\sin(y)=x

Differentiate:

\cos(y)\frac{dy}{dx}=1

So

\frac{dy}{dx}=\frac{1}{\cos(y)}

Using \sin^2(y)+\cos^2(y)=1 and \sin(y)=x gives \cos^2(y)=1-x^2. On the principal range for \arcsin, \cos(y)\ge 0, so \cos(y)=\sqrt{1-x^2} and

\frac{dy}{dx}=\frac{1}{\sqrt{1-x^2}}

Worked example 1: chain rule with inverse trig

Differentiate

y=\arctan(5x^2)

Let g(x)=5x^2, so g'(x)=10x. Then

y'=\frac{10x}{1+(5x^2)^2}=\frac{10x}{1+25x^4}

Worked example 2: combining product rule and inverse trig

Differentiate

y=x\arcsin(x)

Product rule:

y'=\arcsin(x)+\frac{x}{\sqrt{1-x^2}}

A note about other inverse trig derivatives

You may also encounter derivatives for functions like \arcsec(x) in some courses or resources. If they appear, be careful with absolute values and domain restrictions.

Exam Focus

• Typical question patterns:
  • Differentiate expressions like \arcsin(g(x)) and \arctan(g(x)) using chain rule.
  • Use implicit differentiation reasoning (sometimes shown, sometimes just applied).
  • Evaluate derivatives at specific points, possibly involving a triangle interpretation.
• Common mistakes:
  • Forgetting the inside derivative g'(x).
  • Dropping the negative sign on \arccos(x).
  • Mishandling square roots and signs by ignoring principal range restrictions.

Logarithmic and Exponential Differentiation

Why logs and exponentials belong in this unit

Logarithms and exponentials often appear as outside functions wrapped around other expressions, so they naturally combine with the chain rule. They also interact with implicit differentiation, especially when you use logs to turn products into sums.

The natural logarithm and its derivative

\frac{d}{dx}[\ln(x)]=\frac{1}{x}

Because negative inputs sometimes appear during algebraic steps, it is also common to use

\frac{d}{dx}[\ln(|x|)]=\frac{1}{x}

For a composite input g(x),

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

Exponentials and their derivatives

\frac{d}{dx}[e^x]=e^x

With a composite exponent g(x),

\frac{d}{dx}[e^{g(x)}]=e^{g(x)}g'(x)

For general base a>0 and a\ne 1,

\frac{d}{dx}[a^x]=a^x\ln(a)

and

\frac{d}{dx}[a^{g(x)}]=a^{g(x)}\ln(a)g'(x)

Derivatives of logarithms with other bases

Use change of base:

\log_a(x)=\frac{\ln(x)}{\ln(a)}

So

\frac{d}{dx}[\log_a(x)]=\frac{1}{x\ln(a)}

and for g(x),

\frac{d}{dx}[\log_a(g(x))]=\frac{g'(x)}{g(x)\ln(a)}

Worked example 1: log with a composite input

Differentiate

y=\ln(3x^2+1)

Then

y'=\frac{6x}{3x^2+1}

Worked example 2: exponential composite

Differentiate

y=e^{\sin(x)}

Then

y'=e^{\sin(x)}\cos(x)

Logarithmic differentiation (a powerful strategy)

Logarithmic differentiation is especially useful for variable exponents or expressions with many factors. Use log properties:

• \ln(ab)=\ln(a)+\ln(b)
• \ln\left(\frac{a}{b}\right)=\ln(a)-\ln(b)
• \ln(a^k)=k\ln(a)

Method:

1. Take \ln of both sides.
2. Expand using log properties.
3. Differentiate implicitly.
4. Solve for y'.

Worked example 3: variable exponent x^x

Differentiate

y=x^x

Take logs:

\ln(y)=x\ln(x)

Differentiate:

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

So

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

Common pitfalls with logs and exponentials

Forgetting the chain rule is the big one, for example treating \frac{d}{dx}[\ln(g(x))] as \frac{1}{g(x)} instead of \frac{g'(x)}{g(x)}. Also watch the extra factor \ln(a) in base-a exponentials, and remember domain restrictions for logs.

Exam Focus

• Typical question patterns:
  • Differentiate expressions involving \ln(g(x)) or e^{g(x)} using chain rule.
  • Use logarithmic differentiation for variable exponents or many products and quotients.
  • Evaluate derivatives at a point, sometimes using given values of g(x) and g'(x).
• Common mistakes:
  • Dropping g'(x) in log or exponential composites.
  • Forgetting the \ln(a) factor in \frac{d}{dx}[a^x].
  • Misusing log rules, especially trying to turn sums into products.

Combining Differentiation Techniques in AP-Style Problems

Why “mixed-rule” problems are the norm

In Unit 3, derivatives stop being isolated skill checks and start looking like realistic combinations: a product where one factor is a composite, an implicit curve with trig and exponentials, or an inverse function derivative that depends on correctly matching inputs. A good strategy is to stop thinking “Which single rule is it?” and instead think “What operations are happening, and in what order?” Then apply the appropriate rules locally.

A practical decision tree (in words)

• If the expression is a sum or difference, differentiate term-by-term.
• If it is a product, use the product rule (unless expanding is clearly easier).
• If it is a quotient, use the quotient rule (or rewrite with negative exponents if simpler).
• If any part is a function inside a function, apply the chain rule there.
• If y is not solved for, use implicit differentiation.
• If you see f^{-1}, consider the inverse derivative formula.

Worked example 1: product rule + chain rule

Differentiate

y=(x^2+1)\cos(3x)

Let u=x^2+1 and v=\cos(3x). Then

y'=u'v+uv'

Compute u'=2x. For v', chain rule gives

v'=-3\sin(3x)

So

y'=2x\cos(3x)-3(x^2+1)\sin(3x)

Worked example 2: implicit differentiation with an exponential

Find \frac{dy}{dx} if

e^{xy}=x+y

Differentiate. For the left side, chain rule plus product rule inside:

\frac{d}{dx}[e^{xy}]=e^{xy}(x\frac{dy}{dx}+y)

Right side:

\frac{d}{dx}[x+y]=1+\frac{dy}{dx}

Set equal and solve:

e^{xy}(x\frac{dy}{dx}+y)=1+\frac{dy}{dx}

\frac{dy}{dx}=\frac{1-ye^{xy}}{xe^{xy}-1}

Worked example 3: derivative of an inverse at a point using a table

Given

• f(1)=4
• f'(1)=-2

Assuming f is one-to-one, find \left(f^{-1}\right)'(4). Since f^{-1}(4)=1,

\left(f^{-1}\right)'(4)=\frac{1}{f'(1)}=-\frac{1}{2}

Interpreting what your derivative means

Chain rule results like f'(g(x))g'(x) mean “outer rate at the current inner value, times inner rate.” Implicit derivatives often produce slopes in terms of both x and y, reflecting that the slope depends on location on the curve. Inverse derivatives being reciprocals matches the idea that inverses swap inputs and outputs.

Quick hints and habits

• When two terms are multiplied together, use the product rule unless it’s easier to multiply it out.
• If you see a function within another function, you will almost certainly have to use the chain rule.
• If there are x and y terms mixed together, you will need implicit differentiation.
• If you’re finding the derivative at a point, plug the point in at a strategic time to avoid extra algebra, but not so early that you prevent yourself from solving for \frac{dy}{dx}.
• When evaluating derivatives at a point, look for simplifications where factors become 1 or 0.
• You can mentally take certain basic derivatives quickly to reduce clutter, but write enough to avoid sign and factor errors.
• If a second derivative is required, simplify the first derivative before you start.

Exam Focus

• Typical question patterns:
  • Differentiate an expression that requires multiple rules (product or quotient plus chain, or implicit plus chain).
  • Evaluate a complicated derivative at a point using given function values to avoid heavy algebra.
  • Use inverse-derivative relationships in a broader context (tables, graphs, or composed expressions).
• Common mistakes:
  • Applying one rule globally when only part of the expression needs it.
  • Algebra errors after correct differentiation, especially when solving for \frac{dy}{dx}.
  • Not using given numerical values efficiently, leading to unnecessary complexity.