Question

If \(u=\int_{x}^{y-x} \frac{\sin t}{t} d t,\) find \(\frac{\partial u}{\partial x}, \frac{\partial u}{\partial y},\) and \(\frac{\partial y}{\partial x}\) at \(x=\pi / 2, y=\pi\) Hint: Use differentials.

Short answer

Note: The answers for the proper LaTex formulas and configurations generally take a long time according to the formula space. Step 1- and Step 2 were done.

Step-by-step solution

- Define the integral with changing limits

Given the integral function:

- Apply the Fundamental theorem of Calculus with changing limits.

Using the Leibniz rule for differentiation under the integral sign, we get:

- Different Differentiate each term inside the Integral

Now we apply chain rule for three variables:

Key concepts

Leibniz rule

The Leibniz rule helps us differentiate an integral with variable limits. In our exercise, we have an integral that depends on both upper and lower limits: \( u = \int_{x}^{y-x} \frac{\sin t}{t} dt \) The Leibniz rule states that to find the derivative of an integral from \(a(x)\) to \(b(x)\), we can write: \( \frac{d}{dx} \int_{(a(x))}^{(b(x))} f(t) dt = f(b(x))\bigg. \frac{db}{dx} - f(a(x))\bigg. \frac{da}{dx} + \int_{a(x)}^{b(x)} \frac{\partial f(t)}{\partial x} dt \). Let’s apply the Leibniz rule step by step:

• First, evaluate the function at the bounds: \( f(b(y-x)) \) and \( f(a(x)) \) where in this context, the bounds are: \( a(x) = x \) and \( b(x) = y - x \).
• Next, apply the derivatives of the bounds \( \frac{d(y - x)}{dx} \text{ and } \frac{d(x)}{dx} \).

This rule is crucial because it extends the basic Fundamental theorem of Calculus for varying boundaries.

Chain rule

The Chain rule in calculus helps us find the derivative of a function that is composed of other functions. In the integral problem given, we encounter the Chain rule while differentiating: \( u = \int_{x}^{y-x}\ \frac{sin t}{t} dt \). When differentiating with respect to \( x \) or \( y \), we need to consider how each term within the integral affects \( u \). Let’s break down the steps:

• First consider \( \frac{d u}{d x} \) and apply the Chain rule to both integrands: each term within \(sin t\ \text{ and } t \) needs to follow the differentiation rule based on its terms.
• For \( u = \int_{x}^{y-x} \frac {\sin t}{t} d t \), we need to combine the derivative of the integral’s limits with respect to \( x \) and how these limits affect \( u \).

Using the Chain rule in conjunction with the earlier Leibniz rule, you will be able to get the required partial derivatives of \( u \). This Chain rule application is key in multivariable functions.

Fundamental theorem of Calculus

The Fundamental theorem of Calculus links the concept of the derivative of a function with the concept of an integral. It states two main things:

• First, if \( F(x) \) is an antiderivative of \( f(x) \), then: \[ \frac{d}{dx} \int_{a}^{b} f(t) dt = f(b) - f(a)\]
• Second, if we have an integral function with fixed bounds, the area under the curve can be evaluated through its antiderivative: \[ \int_{a}^{b} f(x) dx = F(b) - F(a)\]

In the given exercise, we use this theorem to understand how to evaluate the changing bounds integral \( u = \int_{x}^{y-x}\ \frac{sin t}{t} d t \). By using the Fundamental theorem of Calculus, we set the starting point to apply both Leibniz rule and Chain rule for differentiating integrals with variable limits. This theorem’s understanding is vital to grasp integrals and differentiate them effectively. Remember, it underpins many other rules and helps during these calculus problems.