5.2.4 Calculus Involving Inverse Trig

Differentiating Inverse Trigonometric Functions

Inverse trigonometric functions, such as $\arcsin x$ , $\arccos x$ , and $\arctan x$ , are commonly used in calculus. Differentiating these functions requires understanding their derivatives, which can be derived using implicit differentiation or directly memorized as formulas.

Key Differentiation Rules

Derivative of $\arcsin$ $x$ :

\frac{d}{dx} (\arcsin x) = \frac{1}{\sqrt{1-x^2}}, \quad |x| \leq 1

Derivative of $\arccos$ $x$ :

\frac{d}{dx} (\arccos x) = -\frac{1}{\sqrt{1-x^2}}, \quad |x| \leq 1

Derivative of $\arctan$ $x$ :

\frac{d}{dx} (\arctan x) = \frac{1}{1+x^2}, \quad x \in \mathbb{R}

Derivative of $\text{arccot}\ x$ :

\frac{d}{dx} (\text{arccot}\ x) = -\frac{1}{1+x^2}, \quad x \in \mathbb{R}

Derivative of $\text{arcsec}\ x$ :

\frac{d}{dx} (\text{arcsec}\ x) = \frac{1}{|x| \sqrt{x^2-1}}, \quad |x| > 1

Derivative of $\text{arccsc}\ x$ :

\frac{d}{dx} (\text{arccsc}\ x) = -\frac{1}{|x| \sqrt{x^2-1}}, \quad |x| > 1

Worked Examples

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Example 1: Differentiate $f(x) = \arcsin x + x\sqrt{1-x^2}$

Step 1: Differentiate term by term:

First term:

\frac{d}{dx} (\arcsin x) = \frac{1}{\sqrt{1-x^2}}

Second term:

Use the product rule for $x\sqrt{1-x^2}$

Let $u=x$ and $v = \sqrt{1-x^2}$ :

\frac{d}{dx} (x\sqrt{1-x^2}) = \frac{du}{dx} \times v + u \times \frac{dv}{dx}

\frac{du}{dx} = 1

v = (1-x^2)^{1/2} \implies \frac{dv}{dx} = \frac{-x}{\sqrt{1-x^2}}

Substitute:

\frac{d}{dx} (x\sqrt{1-x^2}) = 1 \times \sqrt{1-x^2} + x \times \frac{-x}{\sqrt{1-x^2}}

= \sqrt{1-x^2} - \frac{x^2}{\sqrt{1-x^2}}

Simplify:

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

Step 2: Combine results:

\frac{d}{dx} \left( \arcsin x + x\sqrt{1-x^2} \right) = \frac{1}{\sqrt{1-x^2}} + \sqrt{1-x^2}

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Example 2**:** Differentiate $f(x) = \frac{1}{2} \arctan(x^2)$

Step 1: Apply the chain rule:

Let $u = x^2$

Then $f(x) = \frac{1}{2} \arctan u$ and $\frac{du}{dx} = 2x$

Step 2: Differentiate $\arctan u$ :

\frac{d}{dx} (\arctan u) = \frac{1}{1+u^2} \times \frac{du}{dx}

Step 3: Substitute back:

\frac{d}{dx} \left( \frac{1}{2} \arctan(x^2) \right) = \frac{1}{2} \times \frac{1}{1+(x^2)^2} \times 2x

= \frac{x}{1+x^4}

Result:

\frac{d}{dx} \left( \frac{1}{2} \arctan(x^2) \right) = \frac{x}{1+x^4}

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Example 3: Differentiate $f(x) = x \arccos x$

Step 1: Use the product rule:

Let $u = x$ and $v = \arccos x$

\frac{d}{dx} (x \arccos x) = \frac{du}{dx} \times v + u \times \frac{dv}{dx}

Step 2: Differentiate $u$ and $v$ :

\frac{du}{dx} = 1

\frac{dv}{dx} = -\frac{1}{\sqrt{1-x^2}}

Step 3: Substitute back:

\frac{d}{dx} (x \arccos x) = 1 \times \arccos x + x \times \left(-\frac{1}{\sqrt{1-x^2}}\right)

Step 4: Simplify:

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

Note Summary

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Common Mistakes:

Mixing up derivatives of $\arcsin x$ and $\arccos x$ : Remember $\arcsin x$ has a positive derivative, whereas $\arccos x$ has a negative one.
Forgetting the chain rule: When differentiating compositions like $\arctan(x^2)$ , always apply the chain rule.
Mismanaging square roots: Be careful with expressions like $\sqrt{1-x^2}$ , especially when simplifying.

Ignoring domain restrictions: Inverse trigonometric functions have specific domains,

📑e.g., $|x| \leq 1$ for $\arcsin x$ and $\arccos x$ .

Omitting the product rule: Don't forget to apply the product rule for terms like $x \arccos x$

infoNote

Key Formulas:

Basic Derivatives:

\frac{d}{dx} (\arcsin x) = \frac{1}{\sqrt{1-x^2}}, \quad \frac{d}{dx} (\arccos x) = -\frac{1}{\sqrt{1-x^2}}

\frac{d}{dx} (\arctan x) = \frac{1}{1+x^2}, \quad \frac{d}{dx} (\text{arccot}\ x) = -\frac{1}{1+x^2}

\frac{d}{dx} (\text{arcsec}\ x) = \frac{1}{|x| \sqrt{x^2-1}}, \quad \frac{d}{dx} (\text{arccsc}\ x) = -\frac{1}{|x| \sqrt{x^2-1}}

Product Rule:

\frac{d}{dx} (u \times v) = \frac{du}{dx} \times v + u \times \frac{dv}{dx}

Chain Rule:

\frac{d}{dx} f(g(x)) = f'(g(x)) \times g'(x)

Calculus Involving Inverse Trig Simplified Revision Notes for A-Level Edexcel Further Maths Core Pure

5.2.4 Calculus Involving Inverse Trig

Differentiating Inverse Trigonometric Functions

Key Differentiation Rules

Derivative of $\arcsin$ $x$ :

Derivative of $\arccos$ $x$ :

Derivative of $\arctan$ $x$ :

Derivative of $\text{arccot}\ x$ :

Derivative of $\text{arcsec}\ x$ :

Derivative of $\text{arccsc}\ x$ :

Worked Examples

Note Summary

Common Mistakes:

Key Formulas:

500K+ Students Use These Powerful Tools to Master Calculus Involving Inverse Trig For their A-Level Exams.

Other Revision Notes related to Calculus Involving Inverse Trig you should explore

Calculus Involving Inverse Trig Simplified Revision Notes for A-Level Edexcel Further Maths Core Pure

5.2.4 Calculus Involving Inverse Trig

Differentiating Inverse Trigonometric Functions

Key Differentiation Rules

Derivative of arcsin⁡\arcsinarcsin xxx:

Derivative of arccos⁡\arccosarccos xxx:

Derivative of arctan⁡\arctanarctan xxx:

Derivative of arccot x\text{arccot}\ xarccot x:

Derivative of arcsec x\text{arcsec}\ xarcsec x:

Derivative of arccsc x\text{arccsc}\ xarccsc x:

Worked Examples

Note Summary

Common Mistakes:

Key Formulas:

500K+ Students Use These Powerful Tools to Master Calculus Involving Inverse Trig For their A-Level Exams.

Other Revision Notes related to Calculus Involving Inverse Trig you should explore

Derivative of $\arcsin$ $x$ :

Derivative of $\arccos$ $x$ :

Derivative of $\arctan$ $x$ :

Derivative of $\text{arccot}\ x$ :

Derivative of $\text{arcsec}\ x$ :

Derivative of $\text{arccsc}\ x$ :