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

Composite function

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

Inside function (inner function)

The function g(x) in f(g(x)); it produces the input to the outside function.

Outside function (outer function)

The function f in f(g(x)); it acts on the output of the inside function.

Chain rule

Differentiation rule for composites: d/dx[f(g(x))] = f'(g(x))·g'(x).

Intermediate variable (u-substitution viewpoint)

Let u=g(x) and y=f(u) to write dy/dx = (dy/du)(du/dx).

“Derivative of the outer, leave the inner”

Chain-rule memory phrase: differentiate the outside function while keeping the inside expression unchanged.

“Multiply by the derivative of the inner”

Final chain-rule step: include an extra factor g'(x) after differentiating the outer function.

“douter, inner, dinner” mnemonic

Mnemonic for chain rule: differentiate the outer, keep the inner, then multiply by the inner’s derivative.

Single-layer composite

A composite with one nesting level (one inside function feeding one outside function).

Multiple-layer composite

A nested composite requiring the chain rule more than once (outer of an inner of another inner).

Inside derivative (g'(x))

The derivative of the inner function; the factor that is commonly forgotten in chain rule problems.

Power composite pattern

If y=(g(x))^n, then y' = n(g(x))^(n−1)·g'(x).

Sine composite pattern

If y=sin(g(x)), then y' = cos(g(x))·g'(x).

Exponential composite pattern (base e)

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

Log composite pattern (natural log)

If y=ln(g(x)), then y' = g'(x)/g(x).

Chain rule evaluation using a table

To find d/dx[f(g(x))] at x=a, compute f'(g(a))·g'(a) using the given values in the correct order.

Common chain rule mistake

Stopping after differentiating the outer function and forgetting to multiply by the inner derivative.

Implicit equation

An equation relating x and y without solving for y as an explicit function of x (e.g., x^2+y^2=25).

Implicit differentiation

Differentiate both sides with respect to x while treating y as y(x), so y-terms require dy/dx via chain rule.

Vertical line test (context)

A test for whether a curve is a function; implicit differentiation is useful for curves that fail it (like circles).

Key implicit idea: y depends on x

Because y=y(x), differentiating any expression involving y must produce a factor of dy/dx.

Derivative of y (implicit)

d/dx[y] = dy/dx.

Derivative of y^2 (implicit)

d/dx[y^2] = 2y·(dy/dx) by chain rule.

Derivative of sin(y) (implicit)

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

Implicit differentiation workflow

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

Product rule in implicit problems

When a term like xy appears, differentiate it using the product rule, not as a single variable.

Derivative of xy (implicit)

d/dx[xy] = x(dy/dx) + y.

Solving for dy/dx

After differentiating implicitly, move all dy/dx terms to one side and isolate dy/dx algebraically.

Reciprocal relationship of derivatives

When defined, dx/dy = 1/(dy/dx); sometimes solving for dx/dy first is cleaner.

Second derivative (implicit)

To find d^2y/dx^2, first compute dy/dx, then differentiate again with respect to x, remembering y still depends on x.

Quotient rule in implicit differentiation

Used when dy/dx is expressed as a quotient; differentiate numerator/denominator carefully, including dy/dx where needed.

Inverse function

A function f^{-1} that undoes f: f(f^{-1}(x))=x and f^{-1}(f(x))=x (on a one-to-one interval).

One-to-one (injective)

A property required for an inverse to be a function; each output corresponds to exactly one input on the interval.

Inverse graph reflection property

Graphs of f and f^{-1} are reflections across the line y=x.

Point-swap rule for inverses

If (a,b) lies on f, then (b,a) lies on f^{-1}.

Derivative of an inverse function (point form)

If f is one-to-one and differentiable with f'(a)≠0, then (f^{-1})'(f(a)) = 1/f'(a).

Derivative of an inverse function (general form)

(f^{-1})'(b) = 1 / f'(f^{-1}(b)).

Condition for inverse derivative to exist

The inverse derivative formula requires f'(a) ≠ 0 at the corresponding point.

Inverse-derivative derivation identity

Start from f(f^{-1}(x))=x; differentiating gives f'(f^{-1}(x))·(f^{-1})'(x)=1.

Notation warning: f^{-1}(x)

f^{-1}(x) means the inverse function, not the reciprocal 1/f(x).

Inverse trigonometric function

A function (like arcsin) that returns an angle whose trig value equals the input, using a restricted principal range.

Principal range

The restricted set of output angles chosen to make an inverse trig function one-to-one.

arcsin(x) principal range

arcsin(x) returns angles in [−π/2, π/2].

arccos(x) principal range

arccos(x) returns angles in [0, π].

arctan(x) principal range

arctan(x) returns angles in (−π/2, π/2).

Derivative of arcsin(x)

d/dx[arcsin(x)] = 1/√(1−x^2).

Derivative of arccos(x)

d/dx[arccos(x)] = −1/√(1−x^2).

Derivative of arctan(x)

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

Chain rule with inverse trig

For u=g(x): d/dx[arcsin(u)] = g'(x)/√(1−u^2) and d/dx[arctan(u)] = g'(x)/(1+u^2).

Logarithmic differentiation

A method for differentiating complicated products/quotients or variable exponents by taking ln of both sides, expanding, differentiating implicitly, then solving for y'.