Calculus I · Unit 2A · lesson

Inverse Trigonometric Derivatives

Concept

Learning objectives

Differentiate inverse trig functions and their compositions.

Inverse Trigonometric Functions

Explanation

Before the formulas

Inverse trigonometric functions reverse restricted trigonometric functions: they accept a ratio and return an angle. Their derivative formulas therefore come from two ingredients used together: implicit differentiation of a relation such as $\sin y=x$ , and a right-triangle identity that rewrites $\cos y$ or $\sin y$ in terms of $x$ .

The branch restriction is part of the function, not a technical footnote. It determines the sign of the radical and explains where a real derivative is finite. State the input domain before applying a formula to a composition.

Explanation

Inverse-trig derivatives come from implicit triangles

Functions such as $\arcsin x$ return angles. Setting $y=\arcsin x$ is equivalent to $\sin y=x$ . Implicit differentiation then produces a derivative involving $\cos y$ , which a right triangle converts back into an expression in $x$ .

This route explains both the square roots and the domain restrictions in the formulas. They are not arbitrary decorations; they encode the geometry of the corresponding inverse relationship.

Inverse trigonometric functions turn ratios into angles. Their derivative formulas contain radicals because the Pythagorean theorem appears when the original trigonometric relation is converted back into algebraic side lengths.

Branch restrictions matter. The symbols $\arcsin$ , $\arccos$ , and $\arctan$ refer to specific inverse branches, not to every angle with the same trigonometric value.

The principal inverse-trigonometric derivatives are

\boxed{\frac{d}{dx}\arcsin x=\frac1{\sqrt{1-x^2}}},\qquad |x|<1,

\boxed{\frac{d}{dx}\arccos x=-\frac1{\sqrt{1-x^2}}},

\boxed{\frac{d}{dx}\arctan x=\frac1{1+x^2}}.

Also,

(\operatorname{arccot}x)'=-\frac1{1+x^2},

(\operatorname{arcsec}x)'=\frac1{|x|\sqrt{x^2-1}}, \quad (\operatorname{arccsc}x)'=-\frac1{|x|\sqrt{x^2-1}}.

Guided walkthrough

Derive the arcsine rule

Let $y=\arcsin x$ , so $\sin y=x$ . Differentiate implicitly:

\cos y\,y'=1.

Thus

y'=\frac1{\cos y}.

On the principal arcsine range, $\cos y\ge0$ , and

\cos y=\sqrt{1-\sin^2y}=\sqrt{1-x^2}.

Therefore

\boxed{\frac{d}{dx}\arcsin x=\frac1{\sqrt{1-x^2}}}.

Worked example

Inverse tangent composition

For $y=\arctan(3x)$ ,

y'=\frac{3}{1+9x^2}.

Worked example

Arcsine of a quadratic

For $y=\arcsin(x^2)$ ,

\boxed{y'=\frac{2x}{\sqrt{1-x^4}}}.

The derivative is real where $|x|<1$ ; endpoint behavior requires one-sided analysis.

After the explanation

Use the section idea

Reading lens

Track which variable depends on which and use reciprocal or logarithmic structure only where its conditions hold.

Mental model

Implicit equations constrain variables together; inverse functions exchange inputs and outputs; logarithms turn products and powers into sums.

Decision

Choose implicit, inverse, or logarithmic differentiation from the equation's representation, not from surface complexity.

Common trap

Dropping a y-prime factor, using a reciprocal slope at the wrong point, or ignoring domain restrictions.

Check yourself

Can you identify the correspondence point and all hidden dependencies before differentiating?

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