Theorem 7.8
Differentiate sinh⁻¹ x = ln (x + √(x² + 1)) to show that d/dx (sinh⁻¹ x) = 1 / √(x² + 1).

검증된 단계별 안내

Start with the given function: \(y = \sinh^{-1} x = \ln \left( x + \sqrt{x^{2} + 1} \right)\).

Differentiate both sides with respect to \(x\) using the chain rule and the derivative of the natural logarithm: \(\frac{dy}{dx} = \frac{1}{x + \sqrt{x^{2} + 1}} \cdot \frac{d}{dx} \left( x + \sqrt{x^{2} + 1} \right)\).

Find the derivative inside the product: \(\frac{d}{dx} \left( x + \sqrt{x^{2} + 1} \right) = 1 + \frac{1}{2 \sqrt{x^{2} + 1}} \cdot 2x = 1 + \frac{x}{\sqrt{x^{2} + 1}}\).

Substitute this back into the expression for \(\frac{dy}{dx}\): \(\frac{dy}{dx} = \frac{1 + \frac{x}{\sqrt{x^{2} + 1}}}{x + \sqrt{x^{2} + 1}}\).

Simplify the expression by combining terms over a common denominator and rationalizing if necessary to show that \(\frac{dy}{dx} = \frac{1}{\sqrt{x^{2} + 1}}\).

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주요 개념

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Inverse Hyperbolic Sine Function (sinh⁻¹ x)

The inverse hyperbolic sine function, sinh⁻¹ x, is defined as the value whose hyperbolic sine is x. It can be expressed using logarithms as sinh⁻¹ x = ln(x + √(x² + 1)), which helps in differentiating it using standard calculus techniques.

Differentiation of Logarithmic Functions

Differentiating a logarithmic function ln(u) involves applying the chain rule: d/dx[ln(u)] = (1/u) * du/dx. This rule is essential for differentiating sinh⁻¹ x when expressed as a logarithm, requiring careful computation of the derivative of the inner function.

Chain Rule and Derivative of Composite Functions

The chain rule allows differentiation of composite functions by multiplying the derivative of the outer function by the derivative of the inner function. In this problem, it is used to differentiate the expression inside the logarithm, particularly the term x + √(x² + 1).

Theorem 7.8Differentiate sinh⁻¹ x = ln (x + √(x² + 1)) to show that d/dx (sinh⁻¹ x) = 1 / √(x² + 1).