Ch. 7 - Logarithmic, Exponential Functions, and Hyperbolic Functions

Briggs - Calculus: Early Transcendentals 3rd Edition

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Briggs 3rd Edition

Ch. 7 - Logarithmic, Exponential Functions, and Hyperbolic Functions

Problem 7.R.18

Chapter 7, Problem 7.R.18

10–19. Derivatives Find the derivatives of the following functions.

g(t) = sinh⁻¹(√t)

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Recall that the function given is \( g(t) = \sinh^{-1}(\sqrt{t}) \), where \( \sinh^{-1}(x) \) is the inverse hyperbolic sine function, also written as \( \text{arsinh}(x) \).

Use the chain rule to differentiate \( g(t) \). The chain rule states that if \( g(t) = f(h(t)) \), then \( g'(t) = f'(h(t)) \cdot h'(t) \). Here, \( f(x) = \sinh^{-1}(x) \) and \( h(t) = \sqrt{t} = t^{1/2} \).

Find the derivative of the outer function \( f(x) = \sinh^{-1}(x) \). The derivative is \( f'(x) = \frac{1}{\sqrt{x^2 + 1}} \).

Find the derivative of the inner function \( h(t) = t^{1/2} \). Using the power rule, \( h'(t) = \frac{1}{2} t^{-1/2} = \frac{1}{2\sqrt{t}} \).

Combine the results using the chain rule: \( g'(t) = f'(h(t)) \cdot h'(t) = \frac{1}{\sqrt{(\sqrt{t})^2 + 1}} \cdot \frac{1}{2\sqrt{t}} \). Simplify the expression inside the square root and write the final derivative expression.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Inverse Hyperbolic Sine Function (sinh⁻¹)

The inverse hyperbolic sine function, sinh⁻¹(x), returns the value whose hyperbolic sine is x. It can be expressed as ln(x + √(x² + 1)), which is useful for differentiation. Understanding its definition helps in applying derivative rules correctly.

Chain Rule

The chain rule is a fundamental differentiation technique used when a function is composed of other functions. It states that the derivative of a composite function is the derivative of the outer function evaluated at the inner function times the derivative of the inner function. This is essential for differentiating g(t) = sinh⁻¹(√t).

Derivative of Square Root Function

The square root function, √t, can be rewritten as t^(1/2). Its derivative is (1/2)t^(-1/2), which is necessary when applying the chain rule to functions involving square roots. Recognizing this derivative simplifies the differentiation process.

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