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

Composite function

A function formed by plugging one function into another, written f(g(x)); apply g first, then f.

Inner function

The inside “layer” of a composite function (often labeled u=g(x)) whose output becomes the input to the outer function.

Outer function

The outside “layer” in a composite function (often y=f(u)) that acts on the output of the inner function.

Layering (in composites)

The idea that one formula can represent multiple nested operations (outside-to-inside structure).

Chain rule

Differentiation rule for compositions: if y=f(g(x)), then dy/dx=f'(g(x))·g'(x).

Leibniz form of the chain rule

Writing dy/dx = (dy/du)·(du/dx) with u=g(x), emphasizing multiplying rates.

“Differentiate outer, keep inner, multiply by inner derivative”

Procedure for chain rule: take derivative of the outer function while leaving the inner expression unchanged, then multiply by the derivative of the inner expression.

“Douter, inner, dinner” mnemonic

Memory trick for chain rule: derivative of outer, keep inner the same, then multiply by derivative of inner.

Derivative notation dy/dx

The derivative of y with respect to x; measures the rate of change of y as x changes.

Prime notation (y′, f′(x))

Alternative derivative notation: y′ means dy/dx; f′(x) is the derivative function of f.

Derivative at a point (f′(a))

The slope of the tangent line to f at x=a (the derivative evaluated at a specific input).

Derivative of a composition notation

Common way to show a chain rule situation: d/dx [f(g(x))].

Common chain rule pitfall

Forgetting to multiply by the derivative of the inner function after differentiating the outer function.

Nested composite function

A composition with multiple layers (e.g., sin(√(1+x^3))) requiring repeated chain rule from outside to inside.

Power rule (as used in chain rule)

When differentiating (u^n), treat u as the variable: d/du[u^n]=n·u^(n−1), then multiply by u′.

Radical-as-power rewrite

Rewriting √(expression) as (expression)^(1/2) to apply the power rule and chain rule more easily.

Radical distribution misconception

The incorrect assumption that √(a+b)=√a+√b; radicals do not distribute over addition.

Units check for chain rule

Interpreting dy/dx as (dy/du)(du/dx); units multiply and cancel to confirm the result’s units make sense.

Product rule

If y=a(x)b(x), then y′=a′(x)b(x)+a(x)b′(x).

Chain rule inside a product

Using chain rule to compute a′(x) or b′(x) when either factor is a composite function.

Common product rule pitfall

Differentiating only one factor or applying product rule but missing the chain rule within a composite factor.

Quotient rule

If y=a(x)/b(x), then y′=[a′(x)b(x)−a(x)b′(x)]/(b(x))^2.

Chain rule inside a quotient

Using chain rule to compute a′(x) and/or b′(x) when numerator or denominator involves composition (like radicals or trig of a function).

Common quotient rule pitfall

Dropping parentheses around a full numerator/denominator term or misplacing the squared denominator.

Function notation composition (h(x)=f(g(x)))

AP-style way to express a composite without formulas; the derivative is h′(x)=f′(g(x))·g′(x).

Table-based chain rule evaluation

Finding h′(a) using given values like g(a), g′(a), and f′(g(a)); evaluate f′ at the inside value.

Common table-based pitfall

Using f′(a) instead of f′(g(a)); the outside derivative must be evaluated at the inner output.

Implicit differentiation

Differentiating an equation relating x and y without solving for y, treating y as a function of x.

Implicitly defined curve

A relation where y is not isolated (e.g., x^2+y^2=25), often representing curves not passing the vertical line test.

Key implicit differentiation idea (y depends on x)

When differentiating with respect to x, treat y as y(x), so derivatives of y-terms include factors of dy/dx.

Chain rule for y^n in implicit differentiation

d/dx[y^n]=n·y^(n−1)·(dy/dx) because y is a function of x.

Implicit differentiation workflow

Differentiate both sides, attach dy/dx to y-terms, collect dy/dx terms on one side, factor, and solve for dy/dx.

Implicit differentiation pitfall (missing dy/dx)

Forgetting dy/dx when differentiating terms like y^2 or sin(y), treating y as if it were constant or equal to x.

Derivative of sin(y) with respect to x

By chain rule: d/dx[sin(y)]=cos(y)·(dy/dx).

Derivative of arctan(y) with respect to x

By chain rule: d/dx[arctan(y)]=(1/(1+y^2))·(dy/dx).

Tangent line (point-slope form)

Equation of the line with slope m through (x1,y1): y−y1=m(x−x1), often using m=dy/dx at that point.

Second derivative (implicit)

d^2y/dx^2 found by differentiating dy/dx again, still treating y as a function of x (y-terms can still trigger chain rule).

Simplify-before-second-derivative tip

Practical strategy: simplify the first derivative expression before differentiating again to reduce algebra errors.

Inverse function

A function f^{-1} that reverses f, so f(f^{-1}(x))=x and f^{-1}(f(x))=x (requires f to be one-to-one on the domain used).

One-to-one (injective)

Property needed for an inverse to exist: each output corresponds to exactly one input (often ensured by restricting the domain).

Reflection across y=x

Geometric relationship: the graph of f^{-1} is the reflection of the graph of f across the line y=x.

Derivative of an inverse function formula

(f^{-1})′(x)=1 / f′(f^{-1}(x)), assuming the inverse exists and f′(f^{-1}(x))≠0.

Inverse derivative at a point

(f^{-1})′(a)=1 / f′(f^{-1}(a)); find b with f(b)=a, then take 1/f′(b).

Inverse function notation misconception

Confusing f^{-1}(x) (inverse function) with 1/f(x) (reciprocal); they are not the same.

Vertical tangent in inverse context

If f′(b)=0, then (f^{-1})′(f(b)) is undefined (inverse has a vertical tangent at the corresponding point).

Inverse trigonometric function ranges

Restricted ranges that make trig inverses one-to-one: arcsin in [−π/2,π/2], arccos in [0,π], arctan in (−π/2,π/2).

Derivative of arcsin(x)

d/dx[arcsin(x)]=1/√(1−x^2) (defined for −1<x<1).

Derivative of arccos(x)

d/dx[arccos(x)]=−1/√(1−x^2) (negative sign is essential).

Derivative of arctan(x)

d/dx[arctan(x)]=1/(1+x^2).

Chain rule with inverse trig (general form)

If y=arcsin(u), then y′=u′/√(1−u^2); if y=arccos(u), then y′=−u′/√(1−u^2); if y=arctan(u), then y′=u′/(1+u^2).