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Ch. 7 - Transcendental Functions
Hass - Thomas' Calculus 15th Edition
Hass15th EditionThomas' CalculusISBN: 9780137616077Not the one you use?Change textbook
All textbooksHass 15th EditionCh. 7 - Transcendental FunctionsProblem 7.3.120
Chapter 7, Problem 7.3.120

In Exercises 115–126, use logarithmic differentiation or the method in Example 6 to find the derivative of y with respect to the given independent variable.
120. y = x^(sin x)

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Recognize that the function is of the form \(y = x^{\sin x}\), where both the base and the exponent depend on \(x\). This suggests using logarithmic differentiation to simplify the differentiation process.
Take the natural logarithm of both sides to get \(\ln y = \ln \left( x^{\sin x} \right)\). Using the logarithm power rule, rewrite this as \(\ln y = \sin x \cdot \ln x\).
Differentiate both sides with respect to \(x\). On the left side, use implicit differentiation: \(\frac{1}{y} \frac{dy}{dx}\). On the right side, apply the product rule to \(\sin x \cdot \ln x\).
Apply the product rule: \(\frac{d}{dx} (\sin x \cdot \ln x) = \cos x \cdot \ln x + \sin x \cdot \frac{1}{x}\).
Solve for \(\frac{dy}{dx}\) by multiplying both sides by \(y\), and then substitute back \(y = x^{\sin x}\) to express the derivative in terms of \(x\).

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Key Concepts

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

Logarithmic Differentiation

Logarithmic differentiation is a technique used to differentiate functions where the variable appears both in the base and the exponent, such as y = x^(sin x). By taking the natural logarithm of both sides, the expression simplifies, allowing the use of implicit differentiation to find the derivative more easily.
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Logarithmic Differentiation

Chain Rule

The chain rule is a fundamental differentiation rule used when differentiating composite functions. In this problem, it is essential for differentiating expressions like sin(x) and the logarithm of x, as these functions are nested within each other after applying logarithmic differentiation.
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Intro to the Chain Rule

Implicit Differentiation

Implicit differentiation involves differentiating both sides of an equation with respect to the independent variable when y is defined implicitly. After taking the logarithm of y, implicit differentiation helps find dy/dx by treating y as a function of x and differentiating accordingly.
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Finding The Implicit Derivative
Related Practice
Textbook Question

Evaluate the integrals in Exercises 87–96.

95. ∫₂⁴ x^(2x) (1 + ln x) dx

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Textbook Question

In Exercises 13–24, find the derivative of y with respect to the appropriate variable.

15. y = 2√t tanh(√t)

Textbook Question

Evaluate the integrals in Exercises 41–60.

49. ∫(sech(√t)tanh(√t)dt)/√t

Textbook Question

13. When is a polynomial f(x) of at most the order of a polynomial g(x) as x→∞? Give reasons for your answer.

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Textbook Question

Since the hyperbolic functions can be expressed in terms of exponential functions, it is possible to express the inverse hyperbolic functions in terms of logarithms, as shown in the following table.

sinh⁻¹x = ln(x + √(x² + 1)), -∞ < x < ∞

cosh⁻¹x = ln(x + √(x² - 1)), x ≥ 1

tanh⁻¹x = (1/2)ln((1+x)/(1-x)), |x| < 1

sech⁻¹x = ln((1+√(1-x²))/x), 0 < x ≤ 1

csch⁻¹x = ln(1/x + √(1+x²)/|x|), x ≠ 1

coth⁻¹x = (1/2)ln((x+1)/(x-1)), |x| > 1

Use these formulas to express the numbers in Exercises 61–66 in terms of natural logarithms.


65. sech⁻¹(3/5)

Textbook Question

In Exercises 139–142, find the length of each curve.

141. y = ln(cos(x)) from x = 0 to x = π/4.

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