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Chapter 4: Problem 10

Find \(d y / d x\) explicitly if \(y=\int_{0}^{1} \frac{e^{x u}-1}{u} d u.\)

Short Answer

Expert verified
The derivative \( \frac{d y}{d x} \) is \( \frac{e^x - 1}{x} \)

Step by step solution

01

- Understand the problem

We're given a function defined by an integral and need to find its derivative with respect to x. The function is: \[ y = \int_{0}^{1} \frac{e^{x u}-1}{u} d u \]
02

- Identify the method

Notice this is an integral with an integrand that depends on both x and u. Use the Leibniz rule for differentiating under the integral sign: \[ \frac{d}{dx} \left( \int_{a(x)}^{b(x)} f(x, u) \, du \right) = \int_{a(x)}^{b(x)} \frac{\partial}{\partial x} f(x, u) \, du + f(x, b(x)) \frac{d}{dx} b(x) - f(x, a(x)) \frac{d}{dx} a(x) \] For this problem, the limits of integration are constants (0 to 1), so the formula simplifies to: \[ \frac{d}{dx} \left( \int_{0}^{1} f(x, u) \, du \right) = \int_{0}^{1} \frac{\partial}{\partial x} f(x, u) \, du \]
03

- Apply the Leibniz rule

The function inside the integral is: \[ f(x, u) = \frac{e^{x u} - 1}{u} \] Calculate the partial derivative with respect to x: \[ \frac{\partial}{\partial x} \left( \frac{e^{x u} - 1}{u} \right) = e^{x u} \]
04

- Integrate the resulting expression

Integrate the partial derivative over the given limits: \[ \frac{d y}{d x} = \int_{0}^{1} e^{x u} \, du \] Solve this integral: \[ \int_{0}^{1} e^{x u} \, du = \left[ \frac{e^{x u}}{x} \right]_{0}^{1} = \frac{e^x - 1}{x} \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Leibniz rule
The Leibniz rule is a powerful tool used in calculus to differentiate an integral whose integrand depends on a parameter. In simpler terms, it allows us to find the derivative of an integral where the integrand (the function inside the integral) depends on more than one variable. For the basic form, if we have an integral '}}

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Most popular questions from this chapter

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Given \(L(q, \dot{q})\) such that \(d L=\dot{p} d q+p d \dot{q},\) find \(H(p, q)\) so that \(d H=\dot{q} d p-\dot{p} d q\) Comments: \(L\) and \(H\) are functions used in mechanics called the Lagrangian and the Hamiltonian. The quantities \(\dot{q}\) and \(\bar{p}\) are actually time derivatives of \(p\) and \(q\), but you make no use of the fact in this problem. Treat \(\dot{p}\) and \(\dot{q}\) as if they were two more variables having nothing to do with \(p\) and \(q\). Hint. Use a Legendre transformation. On your first try you will probably get \(-H .\) Look at the text discussion of Legendre transformations and satisfy yourself that \(g=q y-f\) would have been just as satisfactory as \(g=f-q y\) in (11.23).
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