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

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50 Terms

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Composite function

A function formed by plugging one function into another; if f and g are functions, then (f∘g)(x)=f(g(x)).

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Composition notation (f ∘ g)(x)

The notation for the composite function defined by applying gg first and then ff: (fg)(x)=f(g(x))(f\circ g)(x)=f(g(x)).

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Inner function

In a composite f(g(x)), the inner function is g(x); it is evaluated first and then fed into the outer function.

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Outer function

In a composite f(g(x))f(g(x)), the outer function is ff; it acts on the output of the inner function g(x)g(x).

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Chain Rule

The differentiation rule for composites: ddx[f(g(x))]=f(g(x))g(x)\frac{d}{dx}[f(g(x))]=f'(g(x))\cdot g'(x).

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“Outside evaluated at inside” idea

A Chain Rule phrasing: take the derivative of the outside function, evaluate it at the inside expression, then multiply by the derivative of the inside.

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Leibniz notation Chain Rule

A Chain Rule form using rates: dy/dx=(dy/du)·(du/dx), emphasizing “a rate multiplied by a rate.”

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Derivative operator (d/dx)

An operator meaning “differentiate with respect to x,” as in d/dx[f(x)].

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Prime notation (y')

A common notation for the derivative dy/dx; for example, y'=dy/dx.

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Correct evaluation in Chain Rule

In d/dx[f(g(x))], the outer derivative must be evaluated at g(x), not at x (i.e., f'(g(x))·g'(x), not f'(x)·g'(x)).

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Power-chain situation

A composite where an expression is raised to a power, e.g., (3x2+1)5(3x^2+1)^5; differentiate the power as the outer layer and multiply by the derivative of the inside.

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Chain Rule mistake: forgetting the inside derivative

A common error where students differentiate the outside but do not multiply by g(x)g'(x), the derivative of the inside.

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Chain Rule mistake: losing the inside in trig

A common error like writing cos(x)\cos(x) instead of cos(x2)\cos(x^2) when differentiating sin(x2)\sin(x^2); the trig derivative must keep the full inside expression.

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Multiple-layer Chain Rule

Applying the Chain Rule more than once when there are several nested layers (e.g., e1+xe^{\sqrt{1+x}}).

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Implicit differentiation

Differentiating both sides of an equation involving x and y without solving for y, treating y as y(x) and using the Chain Rule on y-terms.

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Explicit function

A relationship where yy is isolated as a function of xx, e.g., y=x2+3xy=x^2+3x.

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Implicit relationship

A relationship where xx and yy are mixed in an equation and yy is not isolated (e.g., x2+y2=25x^2+y^2=25).

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y depends on x (y(x))

In implicit differentiation, y is treated as a function of x, so differentiating y-expressions requires a dy/dx factor.

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Derivative of y2y^2 with respect to xx

Because y=y(x)y=y(x), ddx[y2]=2ydydx\frac{d}{dx}[y^2]=2y\cdot \frac{dy}{dx} by the Chain Rule.

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Reciprocal derivative relationship

The relationship dx/dy = 1/(dy/dx), used sometimes to switch perspectives between x as a function of y and y as a function of x.

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Implicit differentiation workflow

Differentiate both sides w.r.t. x, include dy/dx on y-terms, gather dy/dx terms on one side, factor, and solve for dy/dx.

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Collect dy/dx terms

A key step in implicit differentiation: move all terms containing dy/dx to one side so you can factor out dy/dx and isolate it.

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Tangent line (general idea)

A line touching a curve at a point with slope equal to the derivative evaluated at that point.

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Point-slope form

A tangent-line form: yy1=m(xx1)y-y_1 = m(x-x_1), where mm is the slope at (x1,y1)(x_1,y_1).

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Slope from implicit differentiation

After finding dy/dx implicitly, substitute the point’s x and y values into dy/dx to get the tangent slope at that point.

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Second derivative implicitly

To find d2ydx2\frac{d^2y}{dx^2}, first compute dydx\frac{dy}{dx}, then differentiate that expression again w.r.t. xx while still treating yy as y(x)y(x).

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Inverse function

A function that reverses another function; if f is one-to-one, then f^{-1} satisfies f(f^{-1}(x))=x and f^{-1}(f(x))=x.

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One-to-one function

A function that passes the horizontal line test on an interval, ensuring it has an inverse function on that interval.

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

For u=g(x)u=g(x), ddx[arcsin(u)]=u1u2\frac{d}{dx}[arcsin(u)]=\frac{u'}{\sqrt{1-u^2}} (and similarly for other inverse trig functions).

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Inverse is not reciprocal

A warning: f^{-1}(x) is not the same as 1/f(x) except in special cases.

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Inverse derivative formula

A general rule: (f^{-1})'(x)=1 / f'(f^{-1}(x)).

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Point-specific inverse derivative formula

If f(a)=b (so f^{-1}(b)=a), then (f^{-1})'(b)=1/f'(a).

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Condition f'(a) ≠ 0 for inverse derivative

For the inverse derivative to exist nicely at a point, ff must be differentiable at aa and have nonzero slope there; otherwise the inverse has a vertical tangent.

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Reflection across y=x (inverse graphs)

The graph of y=f1(x)y=f^{-1}(x) is the reflection of y=f(x)y=f(x) across y=xy=x, and corresponding slopes are reciprocals.

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Principal-value range

A restricted output range used to make inverse trig functions true functions (so trig functions become one-to-one on those intervals).

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arcsin(x) principal range

For y=arcsin(x)y=arcsin(x), yy is restricted to π2yπ2-\frac{\pi}{2} ≤ y ≤ \frac{\pi}{2}.

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arccos(x) principal range

For y=arccos(x)y=arccos(x), yy is restricted to 0yπ0 ≤ y ≤ \pi.

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arctan(x) principal range

For y=arctan(x)y=\arctan(x), y is restricted to π2<y<π2-\frac{\pi}{2} < y < \frac{\pi}{2}.

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Derivative of arcsin(x)

ddx[arcsin(x)]=11x2\frac{d}{dx}[arcsin(x)]=\frac{1}{\sqrt{1-x^2}}.

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Derivative of arccos(x)

ddx[arccos(x)]=11x2\frac{d}{dx}[arccos(x)]=-\frac{1}{\sqrt{1-x^2}}.

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Derivative of arctan(x)

ddx[arctan(x)]=11+x2\frac{d}{dx}[arctan(x)]=\frac{1}{1+x^2}.

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Derivative of arcsec(x)

ddx[arcsec(x)]=1xx21\frac{d}{dx}[arcsec(x)]=\frac{1}{|x|\sqrt{x^2-1}}; the absolute value is important.

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Derivative of arccsc(x)

ddx[arccsc(x)]=1xx21\frac{d}{dx}[arccsc(x)]=-\frac{1}{|x|\sqrt{x^2-1}}; the absolute value is important.

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Chain Rule with inverse trig

A method that takes lnln of both sides to simplify products/quotients/powers (especially when both base and exponent involve xx), then differentiates implicitly.

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Logarithmic differentiation

A method that takes ln of both sides to simplify products/quotients/powers (especially when both base and exponent involve x), then differentiates implicitly.

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Log properties for simplification

Rules used in log differentiation: ln(ab)=ln a+ln b, ln(a/b)=ln a−ln b, and ln(a^k)=k ln a.

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Derivative of ln(y) in log differentiation

Because y=y(x), d/dx[ln(y)]=(1/y)·(dy/dx) by the Chain Rule.

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Derivative of x^x (via log differentiation)

If y=x^x, then y'=x^x(ln x + 1).

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Outermost operation strategy

A problem-solving habit: identify the last operation applied (the “outside”) and differentiate it first, then multiply by the derivative of the inside.

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Log mistake: ln(a+b) is not ln a + ln b

A common error in logarithmic differentiation: logarithms do not distribute over addition, so ln(a+b) ≠ ln a + ln b.