Inverse Trigonometric Derivatives

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\(\dfrac{d}{dx}\arcsin u = \dfrac{u'}{\sqrt{1-u^2}}\)
\(\dfrac{d}{dx}\arccos u = -\dfrac{u'}{\sqrt{1-u^2}}\)
\(\dfrac{d}{dx}\arctan u = \dfrac{u'}{1+u^2}\)

What to Remember

Three derivatives to memorize

\(\dfrac{d}{dx}\arcsin u = \dfrac{u'}{\sqrt{1 - u^2}}\)

\(\dfrac{d}{dx}\arccos u = -\dfrac{u'}{\sqrt{1 - u^2}}\)

\(\dfrac{d}{dx}\arctan u = \dfrac{u'}{1 + u^2}\)

The chain rule supplies u'

If the inside is not \(x\), multiply by its derivative. Leave \(u\) inside the formula until you substitute it.

Arcsine and arccosine are opposites

\(\arcsin x + \arccos x = \dfrac{\pi}{2}\)

Their derivatives add to 0.

Simplify the square root when u is a fraction

For \(\arcsin\!\bigl(\dfrac{x}{2}\bigr)\), \(1 - u^2 = \dfrac{4 - x^2}{4}\). The square root of a quotient becomes a quotient of square roots.

Level 1

Easy

The inside is x, or a constant times x.

Example

Find \(f'(x)\) for \(f(x) = \arcsin x\).

\(\dfrac{d}{dx}\arcsin u = \dfrac{u'}{\sqrt{1 - u^2}}\)

\(u = x\)

\(u' = 1\)

\(\dfrac{1}{\sqrt{1 - x^2}}\)

Answer: \(\dfrac{1}{\sqrt{1 - x^2}}\)

Find \(f'(x)\) for \(f(x) = \arctan x\).

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\(\dfrac{d}{dx}\arctan u = \dfrac{u'}{1 + u^2}\)

\(u = x\)

\(\dfrac{1}{1 + x^2}\)

Answer: \(\dfrac{1}{1 + x^2}\)

Find \(f'(x)\) for \(f(x) = \arccos x\).

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\(\dfrac{d}{dx}\arccos x = -\dfrac{1}{\sqrt{1 - x^2}}\)

Answer: \(-\dfrac{1}{\sqrt{1 - x^2}}\)

Find \(f'(x)\) for \(f(x) = \arctan(2x)\).

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\(u = 2x\)

\(u' = 2\)

\(\dfrac{2}{1 + (2x)^2}\)

\(\dfrac{2}{1 + 4x^2}\)

Answer: \(\dfrac{2}{1 + 4x^2}\)

Find \(f'(x)\) for \(f(x) = \arcsin\!\bigl(\dfrac{x}{2}\bigr)\).

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\(u = \dfrac{x}{2}\)

\(u' = \dfrac{1}{2}\)

\(\dfrac{\dfrac{1}{2}}{\sqrt{1 - \dfrac{x^2}{4}}}\)

\(\dfrac{\dfrac{1}{2}}{\sqrt{\dfrac{4 - x^2}{4}}}\)

\(\dfrac{1}{\sqrt{4 - x^2}}\)

Answer: \(\dfrac{1}{\sqrt{4 - x^2}}\)

Find \(f'(0)\) for \(f(x) = \arcsin(2x)\).

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\(f'(x) = \dfrac{2}{\sqrt{1 - 4x^2}}\)

Substitute \(x = 0\).

\(\dfrac{2}{1}\)

\(2\)

Answer: \(2\)

Level 2

Medium

The inside is a power, an exponential, or a product.

Example

Find \(f'(x)\) for \(f(x) = \arctan(x^2)\).

\(u = x^2\)

\(u' = 2x\)

\(\dfrac{2x}{1 + (x^2)^2}\)

\(\dfrac{2x}{1 + x^4}\)

Answer: \(\dfrac{2x}{1 + x^4}\)

Find \(f'(x)\) for \(f(x) = \arcsin(3x)\).

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\(u = 3x\)

\(u' = 3\)

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

Answer: \(\dfrac{3}{\sqrt{1 - 9x^2}}\)

Find \(f'(x)\) for \(f(x) = \arccos\!\bigl(e^x\bigr)\).

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\(u = e^x\)

\(u' = e^x\)

\(-\dfrac{e^x}{\sqrt{1 - e^{2x}}}\)

Answer: \(-\dfrac{e^x}{\sqrt{1 - e^{2x}}}\)

Find \(f'(x)\) for \(f(x) = x\arctan x\).

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Product rule.

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

\(\arctan x + \dfrac{x}{1 + x^2}\)

Answer: \(\arctan x + \dfrac{x}{1 + x^2}\)

Find \(f'(0)\) for \(f(x) = \arctan(\sin x)\).

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\(f'(x) = \dfrac{\cos x}{1 + \sin^2 x}\)

Substitute \(x = 0\).

\(\dfrac{1}{1 + 0}\)

\(1\)

Answer: \(1\)

Find \(f'(x)\) for \(f(x) = \arcsin x + \arccos x\).

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\(\dfrac{1}{\sqrt{1 - x^2}} + \left(-\dfrac{1}{\sqrt{1 - x^2}}\right)\)

\(0\)

Answer: \(0\)

Level 3

Hard

Simplify a root in the denominator, or evaluate at a point inside the domain.

Example

Find \(f'(x)\) for \(f(x) = \arctan\!\bigl(\sqrt{x}\bigr)\), where \(x > 0\).

\(u = x^{1/2}\)

\(u' = \dfrac{1}{2\sqrt{x}}\)

\(\dfrac{1/(2\sqrt{x})}{1 + x}\)

\(\dfrac{1}{2\sqrt{x}\,(1 + x)}\)

Answer: \(\dfrac{1}{2\sqrt{x}\,(1 + x)}\)

Find \(f'(x)\) for \(f(x) = \ln(\arctan x)\), where \(\arctan x > 0\).

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

\(\dfrac{1}{(1 + x^2)\arctan x}\)

Answer: \(\dfrac{1}{(1 + x^2)\arctan x}\)

Find \(f'(x)\) for \(f(x) = \arctan\!\bigl(\dfrac{1}{x}\bigr)\), where \(x \neq 0\).

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\(u = x^{-1}\)

\(u' = -x^{-2}\)

\(1 + u^2 = \dfrac{x^2 + 1}{x^2}\)

\(\dfrac{-\dfrac{1}{x^2}}{\dfrac{x^2 + 1}{x^2}}\)

\(-\dfrac{1}{1 + x^2}\)

Answer: \(-\dfrac{1}{1 + x^2}\)

Find \(f'\!\bigl(\tfrac{1}{4}\bigr)\) for \(f(x) = \arcsin(2x)\).

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\(f'(x) = \dfrac{2}{\sqrt{1 - 4x^2}}\)

Substitute \(x = \tfrac{1}{4}\).

\(\dfrac{2}{\sqrt{1 - 4\cdot\tfrac{1}{16}}}\)

\(\dfrac{2}{\sqrt{\dfrac{3}{4}}}\)

\(\dfrac{2}{\dfrac{\sqrt{3}}{2}}\)

\(\dfrac{4}{\sqrt{3}}\)

\(\dfrac{4\sqrt{3}}{3}\)

Answer: \(\dfrac{4\sqrt{3}}{3}\)

Which expression equals the derivative of \(\arccos(2x)\)?

A. \(-\dfrac{2}{\sqrt{1 - 4x^2}}\)

B. \(\dfrac{2}{\sqrt{1 - 4x^2}}\)

C. \(-\dfrac{1}{\sqrt{1 - 4x^2}}\)

D. \(-\dfrac{2}{1 + 4x^2}\)

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\(u = 2x\)

\(u' = 2\)

\(-\dfrac{2}{\sqrt{1 - 4x^2}}\)

Answer: A. \(-\dfrac{2}{\sqrt{1 - 4x^2}}\)

For \(f(x) = \arctan\!\bigl(e^x\bigr)\), find \(f'(0)\).

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\(u = e^x\)

\(u' = e^x\)

\(f'(x) = \dfrac{e^x}{1 + \bigl(e^x\bigr)^2}\)

\(f'(x) = \dfrac{e^x}{1 + e^{2x}}\)

Substitute \(x = 0\).

\(\dfrac{1}{1 + 1}\)

\(\dfrac{1}{2}\)

Answer: \(\dfrac{1}{2}\)

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