Understanding Derivatives of Inverses (Including Inverse Trig)

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25 Terms

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Inverse function

A function that undoes another function: if y=f(x), then x=f^{-1}(y), so f(f^{-1}(x))=x and f^{-1}(f(x))=x.

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Defining property of inverses

The pair of identities f(f1(x))=xf(f^{-1}(x))=x and f1(f(x))=xf^{-1}(f(x))=x (on appropriate domains), which characterizes inverse functions.

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Reciprocal function (common confusion)

The function 1/f(x); it is NOT the same as f^{-1}(x), which denotes an inverse function.

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Reflection across y=x (inverse graphs)

The graph of y=f1(x)y=f^{-1}(x) is the reflection of the graph of y=f(x)y=f(x) across the line y=xy=x, swapping x- and y-coordinates.

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One-to-one function

A function where each output corresponds to exactly one input (passes the horizontal line test); required for an inverse to be a function.

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Differentiability condition for inverse-derivative rule

To use (f1)(a)=1f(f1(a))(f^{-1})'(a)=\frac{1}{f'(f^{-1}(a))}, ff must be one-to-one and differentiable at the matching point with f(that point)0f'(\text{that point})≠0.

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Inverse-derivative formula

If ff is differentiable and one-to-one and f(f1(x))0f'(f^{-1}(x))\neq 0, then ddx[f1(x)]=1f(f1(x))\frac{d}{dx}[f^{-1}(x)]=\frac{1}{f'(f^{-1}(x))}.

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Point-specific inverse derivative

(f1)(a)=1f(f1(a))(f^{-1})'(a)=\frac{1}{f'(f^{-1}(a))}; find bb such that f(b)=af(b)=a, then compute 1f(b)\frac{1}{f'(b)}.

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Matching points for a function and its inverse

If (b,a)(b,a) lies on y=f(x)y=f(x) (so f(b)=af(b)=a), then (a,b)(a,b) lies on y=f1(x)y=f^{-1}(x); these are the corresponding points where slopes reciprocate.

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Reciprocal slope idea

The slope of the inverse at x=a equals the reciprocal of the slope of the original at x=b where f(b)=a (inputs swap when inverting).

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Vertical tangent (inverse context)

If f(b)=0f'(b)=0, then the inverse has an undefined slope at x=f(b)x=f(b) (a vertical tangent), so (f1)(f(b))(f^{-1})'(f(b)) is not finite/undefined.

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Implicit differentiation (inverse functions)

A method to differentiate an inverse relationship by writing f(g(x))=x and differentiating both sides using the chain rule.

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Inverse trigonometric function

A function (like arcsin, arccos, arctan) that reverses a trig function after restricting the trig function to a one-to-one principal interval.

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Principal range

The restricted output interval used to make an inverse trig function a true function; it determines correct signs (e.g., sign of cos or sin) in derivations.

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arcsin(x)

Inverse sine: y=arcsin(x)y=\text{arcsin}(x) means sin(y)=x\text{sin}(y)=x with yy in the principal range [θ2,θ2][-\frac{\theta}{2}, \frac{\theta}{2}].

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arccos(x)

Inverse cosine: y=arccos(x)y=\text{arccos}(x) means cos(y)=x\text{cos}(y)=x with yy in the principal range [0,π][0, \text{π}].

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arctan(x)

Inverse tangent: y=arctan(x)y=\text{arctan}(x) means tan(y)=x\text{tan}(y)=x with yy in the principal range (π2,π2)(-\frac{\text{π}}{2}, \frac{\text{π}}{2}).

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Domain and range of arcsin(x)

Domain [1,1][-1,1]; range [θ2,θ2][-\frac{\theta}{2}, \frac{\theta}{2}].

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Domain and range of arccos(x)

Domain [1,1][-1,1]; range [0,θ][0, \theta].

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Domain and range of arctan(x)

Domain (,)(-\text{∞},\text{∞}); range (π2,π2)(-\frac{\text{π}}{2}, \frac{\text{π}}{2}).

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Inverse trig notation warning

sin1(x),cos1(x),tan1(x)\text{sin}^{-1}(x), \text{cos}^{-1}(x), \text{tan}^{-1}(x) mean arcsin(x),arccos(x),arctan(x)\text{arcsin}(x), \text{arccos}(x), \text{arctan}(x)—not 1sin(x),1cos(x),1tan(x)\frac{1}{\text{sin}(x)}, \frac{1}{\text{cos}(x)}, \frac{1}{\text{tan}(x)}.

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Derivative of arcsin(x)

$$\frac{d}{dx}[\text{arcsin}(x)]=\frac{1}{\text{√}(1−x^2)} \text{ (for } |x|

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Derivative of arccos(x)

ddx[arccos(x)]=1sqrt(1x2)\frac{d}{dx}[\text{arccos}(x)]=−\frac{1}{\text{sqrt}(1−x^2)} (for x<1|x|<1).

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Derivative of arctan(x)

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

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Chain rule with inverse trig

If u=u(x)u=u(x), then ddx[arcsin(u)]=usqrt(1u2)\frac{d}{dx}[\text{arcsin}(u)]=\frac{u'}{\text{sqrt}(1−u^2)}, ddx[arccos(u)]=usqrt(1u2)\frac{d}{dx}[\text{arccos}(u)]=−\frac{u'}{\text{sqrt}(1−u^2)}, and ddx[arctan(u)]=u(1+u2)\frac{d}{dx}[\text{arctan}(u)]=\frac{u'}{(1+u^2)}.