Problem 89 - Calculus Early Transcendentals

Question

Let \(h\) be continuous for all real numbers. a. Find \(f_{x}\) and \(f_{y}\) when \(f(x, y)=\int_{x}^{y} h(s) d s\) b. Find \(f_{x}\) and \(f_{y}\) when \(f(x, y)=\int_{1}^{x y} h(s) d s\)

Step-by-Step Solution

Answer

In summary, for the given function \(f(x, y)\), the partial derivatives are: Part (a): - \(f_x = -h(x)\) - \(f_y = h(y)\) Part (b): - \(f_x = yh(xy)\) - \(f_y = xh(xy)\)

Step 1: Part (a) - Finding \(f_x\) #

To find \(f_x\), let's apply the Leibniz rule: $$ f_x = \frac{\partial}{\partial x} \int_x^y h(s) ds = h(y) \frac{\partial y}{\partial x} - h(x) \frac{\partial x}{\partial x} $$ Since \(\frac{\partial y}{\partial x} = 0\) and \(\frac{\partial x}{\partial x} = 1\), we have: $$ f_x = -h(x) $$ Note that this result is valid as \(h\) is continuous for all real numbers.

Step 2: Part (a) - Finding \(f_y\) #

To find \(f_y\), we apply the Leibniz rule again: $$ f_y = \frac{\partial}{\partial y}\int_x^y h(s) ds = h(y) \frac{\partial y}{\partial y} - h(x) \frac{\partial x}{\partial y} $$ Since \(\frac{\partial y}{\partial y} = 1\) and \(\frac{\partial x}{\partial y} = 0\), we have: $$ f_y = h(y) $$ The results for part (a) are \(f_x = -h(x)\) and \(f_y = h(y)\).

Step 3: Part (b) - Finding \(f_x\) #

To find \(f_x\), let's apply the Leibniz rule and the product rule: $$ f_x = \frac{\partial}{\partial x} \int_1^{xy} h(s) ds = h(xy) \frac{\partial (xy)}{\partial x} = yh(xy) $$ This result is valid since \(h\) is continuous for all real numbers.

Step 4: Part (b) - Finding \(f_y\) #

Similarly, to find \(f_y\), we apply the Leibniz rule and the product rule: $$ f_y = \frac{\partial}{\partial y} \int_1^{xy} h(s) ds = h(xy) \frac{\partial (xy)}{\partial y} = xh(xy) $$ The results for part (b) are \(f_x = yh(xy)\) and \(f_y = xh(xy)\).

Key Concepts

Leibniz Rule

The Leibniz rule is an essential tool in calculus for differentiating under the integral sign. It allows us to find the derivative of an integral where the limits of integration are functions of the variable we're differentiating with respect to. This might sound complex, but let's break it down.

Consider we have a function defined as an integral:

• \( f(x, y) = \int_{a(x)}^{b(y)} h(s) ds \)

Here, both the upper limit \( b(y) \) and lower limit \( a(x) \) can be functions of \( x \) or \( y \).

To use the Leibniz rule to find \( f_x \) (the derivative with respect to \( x \)), we'd write:

• \( f_x = h(b(y)) \frac{\partial b}{\partial x} - h(a(x)) \frac{\partial a}{\partial x} \)

Similarly, for \( f_y \):

• \( f_y = h(b(y)) \frac{\partial b}{\partial y} - h(a(x)) \frac{\partial a}{\partial y} \)

The approach involves differentiating the limits, which tells us how the boundary of the area under the integral changes. This methodology is critical for problems where limits move, such as those based on dynamic systems.

Integral Calculus

Integral calculus deals with accumulation of quantities and areas under curves. In the context of this exercise, we're interested in definite integrals, which mean evaluating an integral between two points. This is what we've done with the functions \( f(x, y) = \int_{x}^{y} h(s) ds \) and \( f(x, y) = \int_{1}^{xy} h(s) ds \).

When we compute these integrals, we're essentially summing up all the infinitesimal products of \( h(s) \) over the interval from \( x \) to \( y \) or another specified range. The fundamental theorem of calculus helps us link these integrals with derivatives, allowing us to use tools such as the Leibniz rule to explore changes in our integral as functions change.

Integral calculus doesn't just stop at finding the area under curves, it also helps solve differential equations, calculate probabilities, analyze forces in physics, and even find centers of mass. In our exercise, the ability to move from definite integral representation to derivative form is a beautiful illustration of the power of calculus.

Continuity of Functions

Continuity of a function ensures that the function behaves well at all points in its domain. In mathematical terms, we say a function is continuous at a point if the limit of the function as it approaches the point equals the function's value at that point. Why does this matter in calculus?

For our exercise, function \( h(s) \) is assumed continuous. This ensures the integrals we're working with are well-behaved and that tools like the Leibniz rule apply smoothly. If \( h(s) \) weren't continuous, we might face issues like jumps or gaps in behavior, which can make calculus operations tricky or even invalid.

There are several important implications of function continuity:

• It allows the interchange of limits and integration, essential for techniques like the Leibniz rule.
• Ensures that the integral from any two bounds is defined and finite.
• Guarantees the differentiability of the integral in cases described by the fundamental theorem of calculus.

Continuity is like a promise that tells calculus, "Go ahead, I won't throw any surprises at you!" Knowing a function is continuous allows us to apply many calculus operations with confidence.