Derivative of Inverse Trigonometric Functions

You already know how to differentiate \(\sin x\), \(\cos x\), and \(\tan x\) using the rules covered in derivative trigonometry. The inverse trig functions, \(\arcsin x\), \(\arccos x\), \(\arctan x\), and the rest, undo those functions, and their derivatives look completely different: instead of more trig ratios, you get algebraic expressions built from square roots and squares. Recognizing these formulas by sight is essential, because inverse trig derivatives show up constantly in chain rule problems, integration by substitution, and related-rates questions later in calculus.

Before diving in, it helps to be comfortable with how angles and ratios relate in trigonometric ratios in radians, since the proofs below lean on right-triangle relationships between an angle and its sine, cosine, or tangent.

Let \(u\) be a differentiable function of \(x\). The six inverse trigonometric derivative rules are:

Two patterns are worth noticing right away. First, the "co" versions (\(\arccos\), \(\)arccot\(\), \(\)arccsc\(\)) are always the negative of their partner. Second, when \(u\) is itself a function of \(x\) rather than just \(x\), every formula picks up an extra factor of \(u'\) from the chain rule, for example \(\dfrac{d}{dx}\big[\arcsin u\big] = \dfrac{u'}{\sqrt{1-u^2}}\).

These rules are not arbitrary, they follow from implicit differentiation applied to the definition of the inverse function. Let \(y = \arcsin x\). By definition this means \(\sin y = x\), with \(y\) restricted to \(\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]\).

Differentiate both sides of \(\sin y = x\) with respect to \(x\), treating \(y\) as a function of \(x\):

\(\cos y \cdot \dfrac{dy}{dx} = 1\)

Solve for \(\dfrac{dy}{dx}\):

\(\dfrac{dy}{dx} = \dfrac{1}{\cos y}\)

The formula needs to be in terms of \(x\), not \(y\). Since \(\sin y = x\), picture a right triangle with opposite side \(x\), hypotenuse \(1\), and adjacent side \(\sqrt{1-x^2}\) (from the Pythagorean identity \(\sin^2 y + \cos^2 y = 1\)). That gives \(\cos y = \sqrt{1-x^2}\), so:

\(\dfrac{d}{dx}\big[\arcsin x\big] = \dfrac{1}{\sqrt{1-x^2}}\)

The same technique, starting from \(\cos y = x\) or \(\tan y = x\), produces every other formula in the table. Once you have seen one proof, you have effectively seen all six.

Looking at the graphs side by side makes the algebra concrete. \(\arctan x\) is an increasing, S-shaped curve that flattens out as \(x\) grows, which matches the fact that its slope, \(\dfrac{1}{1+x^2}\), gets smaller and smaller (but never negative) as \(|x|\) increases.

Differentiate \(f(x) = \arcsin(3x)\).

Here \(u = 3x\), so \(u' = 3\). Using \(\dfrac{d}{dx}\big[\arcsin u\big] = \dfrac{u'}{\sqrt{1-u^2}}\):

\(f'(x) = \dfrac{3}{\sqrt{1-9x^2}}\)

Differentiate \(g(x) = x^2 \arctan x\).

Apply the product rule with \(u = x^2\) and \(v = \arctan x\):

\(g'(x) = 2x \arctan x + x^2 \cdot \dfrac{1}{1+x^2}\)

This simplifies to \(g'(x) = 2x\arctan x + \dfrac{x^2}{1+x^2}\). Notice how this problem still uses the same product rule and chain rule mechanics from finding slope and equation of tangent line problems, only one of the pieces happens to be an inverse trig function.

Differentiate \(h(x) = \dfrac{\arccos x}{x}\).

By the quotient rule, with \(\dfrac{d}{dx}\arccos x = \dfrac{-1}{\sqrt{1-x^2}}\):

\(h'(x) = \dfrac{\dfrac{-x}{\sqrt{1-x^2}} - \arccos x}{x^2}\)

Students mixing up inverse trig derivatives usually fall into one of these traps:

• Confusing \(\arcsin x\) with \(\dfrac{1}{\sin x}\), the two are completely different functions with different derivatives.
• Forgetting the negative sign on the "co" functions: \(\arccos x\), \(\)arccot \( x\), and \(\)arccsc \( x\) all have negative derivatives.
• Dropping the chain rule factor \(u'\) when the inside function is not simply \(x\).
• Losing the absolute value \(|x|\) in the \(\)arcsec\(\) and \(\)arccsc\(\) formulas.

If you ever want to double-check a derivative numerically rather than symbolically, the method used in estimating derivatives from a table is a useful sanity check for any of these formulas at a specific point.

Once the first derivatives of inverse trig functions feel automatic, the next natural steps are taking second and third derivatives, covered in higher order derivatives, and comparing them with the derivative rules for exponential and logarithmic functions, since the inverse of \(e^x\) follows the very same implicit differentiation strategy used above.

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