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 $\frac{dy}{dx}=f'(g(x))\bullet g'(x)$.

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 $\frac{dy}{dx}$; $f′(x)$ 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., $\sin(\sqrt{1+x^3})$) requiring repeated chain rule from outside to inside.

Power rule (as used in chain rule)

When differentiating $u^n$, treat $u$ as the variable: $\frac{d}{du}[u^n]=n\cdot u^{n-1}$, 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 $y=\frac{a(x)}{b(x)}$, then $y′=\frac{[a′(x)b(x)-a(x)b′(x)]}{(b(x))^2}$.

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 $h'(x)=f'(g(x))\bullet g'(x)$.

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

$\frac{d}{dx}[y^n]=n\cdot y^{n-1}\cdot \frac{dy}{dx}$ 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 $\frac{dy}{dx}$ 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 $(x_1,y_1)$: $y-y_1=m(x-x_1)$, often using $m=\frac{dy}{dx}$ at that point.

Second derivative (implicit)

$\frac{d^2y}{dx^2}$ found by differentiating $\frac{dy}{dx}$ 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 $f′(f^{-1}(x)) \ne 0$.

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 $[-\frac{\pi}{2},\frac{\pi}{2}]$, $\arccos$ in $[0,\pi]$, $\arctan$ in $(-\frac{\pi}{2},\frac{\pi}{2})$.

Derivative of arcsin(x)

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

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

Chain rule with inverse trig (general form)

If $y=\arcsin(u)$, then $y′=\frac{u′}{\sqrt{1-u^2}}$; if $y=\arccos(u)$, then $y′=-\frac{u′}{\sqrt{1-u^2}}$; if $y=\arctan(u)$, then $y′=\frac{u′}{1+u^2}$.