Derivatives of Inverse Trigonometric Functions 3.10

Derivatives of Inverse Trigonometric Functions

Introduction

Inverse trigonometric functions are essential in calculus for solving equations involving trigonometric expressions. Their derivatives are frequently used in integration, differentiation, and applications involving rates of change. This section summarizes the key formulas and provides worked examples for finding derivatives involving inverse trigonometric functions.

Key Derivative Formulas

• Derivative of arcsin (inverse sine): $\frac{d}{dx} (\sin^{-1} x) = \frac{1}{\sqrt{1 - x^2}}$

• Derivative of arccos (inverse cosine): $\frac{d}{dx} (\cos^{-1} x) = -\frac{1}{\sqrt{1 - x^2}}$

• Derivative of arctan (inverse tangent): $\frac{d}{dx} (\tan^{-1} x) = \frac{1}{1 + x^2}$

• Derivative of arccot (inverse cotangent): $\frac{d}{dx} (\cot^{-1} x) = -\frac{1}{1 + x^2}$

• Derivative of arcsec (inverse secant): $\frac{d}{dx} (\sec^{-1} x) = \frac{1}{|x| \sqrt{x^2 - 1}}$

• Derivative of arccsc (inverse cosecant): $\frac{d}{dx} (\csc^{-1} x) = -\frac{1}{|x| \sqrt{x^2 - 1}}$

Examples and Applications

Example 1: Differentiating $y = \sin^{-1}(8x^4)$

• Let $u = 8x^4$.

• By the chain rule: $y' = \frac{1}{\sqrt{1 - (8x^4)^2}} \cdot (8x^4)'$

• Compute the derivative: $y' = \frac{1}{\sqrt{1 - 64x^8}} \cdot 32x^3 = \frac{32x^3}{\sqrt{1 - 64x^8}}$

Example 2: Differentiating $y = \tan^{-1}(e^{3x})$

• Let $u = e^{3x}$.

• By the chain rule: $y' = \frac{1}{1 + (e^{3x})^2} \cdot (e^{3x})'$

• Compute the derivative: $y' = \frac{1}{1 + e^{6x}} \cdot 3e^{3x} = \frac{3e^{3x}}{1 + e^{6x}}$

Example 3: Differentiating $y = (\cos^{-1} t)^2$

• Let $u = \cos^{-1} t$.

• By the chain rule: $y' = 2\cos^{-1} t \cdot \frac{d}{dt}(\cos^{-1} t)$

• Recall $\frac{d}{dt}(\cos^{-1} t) = -\frac{1}{\sqrt{1 - t^2}}$

• So: $y' = 2\cos^{-1} t \cdot \left(-\frac{1}{\sqrt{1 - t^2}}\right) = -\frac{2\cos^{-1} t}{\sqrt{1 - t^2}}$

Example 4: Differentiating $y = 3x \sin^{-1} x$

• Use the product rule:

• $y' = 3 \sin^{-1} x + 3x \cdot \frac{1}{\sqrt{1 - x^2}}$

• So: $y' = 3 \sin^{-1} x + \frac{3x}{\sqrt{1 - x^2}}$

Example 5: Composite Function Derivative and Evaluation

Given: $f(x) = x \tan^{-1}\left(\frac{x}{2}\right)$, find $f'(2)$.

• Apply the product rule:

• $f'(x) = 1 \cdot \tan^{-1}\left(\frac{x}{2}\right) + x \cdot \frac{1}{1 + (\frac{x}{2})^2} \cdot \frac{1}{2}$

• Simplify: $f'(x) = \tan^{-1}\left(\frac{x}{2}\right) + \frac{x}{1 + (x/2)^2} \cdot \frac{1}{2}$

• At $x = 2$: $f'(2) = \tan^{-1}(1) + 2 \cdot \frac{1}{1 + 1} \cdot \frac{1}{2} = \frac{\pi}{4} + \frac{1}{2}$

Example 6: Tangent Line to $g(x) = \sec^{-1}(2x)$ at $(1, \frac{\pi}{3})$

• Derivative formula: $\frac{d}{dx} (\sec^{-1} u) = \frac{1}{|u| \sqrt{u^2 - 1}} \cdot u'$

• For $g(x) = \sec^{-1}(2x)$: $g'(x) = \frac{1}{|2x| \sqrt{(2x)^2 - 1}} \cdot 2 = \frac{1}{|x| \sqrt{4x^2 - 1}}$

• At $x = 1$: $g'(1) = \frac{1}{1 \cdot \sqrt{4 - 1}} = \frac{1}{\sqrt{3}}$

• Equation of tangent line at $(1, \frac{\pi}{3})$: $y - \frac{\pi}{3} = \frac{1}{\sqrt{3}} (x - 1)$ Or, equivalently: $y = \frac{1}{\sqrt{3}} (x - 1) + \frac{\pi}{3}$

Summary Table: Derivatives of Inverse Trigonometric Functions

Function	Derivative
$\sin^{-1} x$	$\frac{1}{\sqrt{1 - x^2}}$
$\cos^{-1} x$	$-\frac{1}{\sqrt{1 - x^2}}$
$\tan^{-1} x$	$\frac{1}{1 + x^2}$
$\cot^{-1} x$	$-\frac{1}{1 + x^2}$
$\sec^{-1} x$	$\frac{1}{|x| \sqrt{x^2 - 1}}$
$\csc^{-1} x$	$-\frac{1}{|x| \sqrt{x^2 - 1}}$

Additional info:

• These derivatives are valid for all $x$ in the domains of the respective inverse trigonometric functions.

• When differentiating composite functions involving inverse trigonometric functions, always apply the chain rule.

• Applications include finding tangent lines, solving integrals, and modeling physical phenomena involving angles and rates.