Ch. 7 - Transcendental Functions

Hass15th EditionThomas' CalculusISBN: 9780137616077Not the one you use?Change textbook

Hass 15th Edition

Ch. 7 - Transcendental Functions

Problem 7.AAE.9

Chapter 7, Problem 7.AAE.9

In Exercises 9 and 10, use implicit differentiation to find dy/dx.
9. y^e^x = x^y + 1

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Start with the given equation: \(y^{e^{x}} = x^{y} + 1\).

Differentiate both sides of the equation with respect to \(x\) using implicit differentiation. Remember that \(y\) is a function of \(x\), so when differentiating terms involving \(y\), apply the chain rule.

For the left side, \(y^{e^{x}}\), use logarithmic differentiation: take the natural logarithm of both sides or apply the formula for differentiating \(u^{v}\) where both \(u\) and \(v\) depend on \(x\). The derivative is \(\frac{d}{dx} y^{e^{x}} = y^{e^{x}} \left( e^{x} \ln(y) + e^{x} \frac{1}{y} \frac{dy}{dx} \right)\).

For the right side, differentiate \(x^{y}\) using implicit differentiation: \(\frac{d}{dx} x^{y} = x^{y} \left( \frac{y}{x} + \ln(x) \frac{dy}{dx} \right)\), and the derivative of the constant \(1\) is \(0\).

After differentiating both sides, collect all terms involving \(\frac{dy}{dx}\) on one side and factor it out. Then solve for \(\frac{dy}{dx}\) to express it explicitly in terms of \(x\), \(y\), and known functions.

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Key Concepts

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

Implicit Differentiation

Implicit differentiation is a technique used to find the derivative of a function when y is defined implicitly rather than explicitly in terms of x. It involves differentiating both sides of the equation with respect to x, treating y as a function of x, and applying the chain rule when differentiating terms involving y.

Chain Rule

The chain rule is a fundamental differentiation rule used when differentiating composite functions. In implicit differentiation, it allows us to differentiate terms involving y by multiplying the derivative of the outer function by dy/dx, since y depends on x.

Logarithmic Differentiation

Logarithmic differentiation is useful when differentiating expressions where variables appear as both bases and exponents, such as y^e^x or x^y. Taking the natural logarithm of both sides simplifies the differentiation process by converting products and powers into sums and products, making implicit differentiation more manageable.

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Logarithmic Differentiation

Related Practice

Textbook Question

20. Solid of revolution The region between the curve y=1/(2√x) and the x-axis from x=1/4 to x=4 is revolved about the x-axis to generate a solid.

a. Find the volume of the solid.

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Textbook Question

Find the areas between the curves y=2(log_2(x))/x and y=2(log_4(x))/x and the x-axis from x=1 to x=e. What is the ratio of the larger area to the smaller?

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Textbook Question

Find the limits in Exercises 1–6.

5. lim(n→∞) (1/(n+1) + 1/(n+2) + ... + 1/(2n))

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Textbook Question

7. Let A(t) be the area of the region in the first quadrant enclosed by the coordinate axes, the curve y=e^(-x), and the vertical line x=t, t>0. Let V(t) be the volume of the solid generated by revolving the region about the x-axis. Find the following limits.

a. lim(x→∞)A(t)

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Textbook Question

7. What integrals lead to logarithms? Give examples. What are the integrals of tan x, cot x, sec x, and csc x?

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Textbook Question

Find the limits in Exercises 1–6.

1. lim(b→1⁻) ∫(from 0 to b) dx/√(1-x²)

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In Exercises 9 and 10, use implicit differentiation to find dy/dx.9. y^e^x = x^y + 1

Verified video answer for a similar problem:

Key Concepts

Implicit Differentiation

Chain Rule

Logarithmic Differentiation

In Exercises 9 and 10, use implicit differentiation to find dy/dx.
9. y^e^x = x^y + 1