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Chapter 15: Problem 99

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

Short Answer

Expert verified
Question: Find the partial derivatives of the function \(f(x,y) = \int_{1}^{xy} h(s) ds \) with respect to \(x\) and \(y\), where \(h(s)\) is a continuous function. Answer: The partial derivatives of the function are: $$f_x = h(xy) y$$ $$f_y = h(xy) x$$

Step by step solution

01

Understand Leibniz Rule

Leibniz Rule states that if we want to differentiate an integral with respect to some variable, we can differentiate inside the integral with respect to that variable and keep the integral sign: $$\frac{d}{dx}\int_{a(x)}^{b(x)} g(x,t) dt = \int_{a(x)}^{b(x)} \frac{d g(x,t)}{dx} dt + g(x, b(x)) b'(x) - g(x,a(x))a'(x)$$ In our case, the function is \(f(x,y) = \int_{1}^{xy} h(s) ds\).
02

Find \(f_x\) (partial derivative with respect to x)

We will differentiate the function \(f(x,y)\) with respect to \(x\) using Leibniz Rule: $$f_x = \frac{\partial}{\partial x}\int_{1}^{xy} h(s) ds $$ From the Leibniz Rule, we have: $$f_x = \int_{1}^{xy} \frac{\partial h(s)}{\partial x} ds + h(xy) \frac{\partial (xy)}{\partial x} - h(1) \frac{\partial 1}{\partial x} $$ Notice that \(\frac{\partial h(s)}{\partial x}\) is zero because there is no \(x\) in \(h(s)\). The last term on the right side vanishes since the derivative of a constant is zero: $$f_x = 0 + h(xy) \frac{\partial (xy)}{\partial x} - 0 $$ Now, we just need to find the derivative of \((xy)\) with respect to \(x\): $$\frac{\partial (xy)}{\partial x} = y$$ So, we have: $$f_x = h(xy) y$$
03

Find \(f_y\) (partial derivative with respect to y)

Now, we will differentiate the function \(f(x,y)\) with respect to \(y\) using Leibniz Rule: $$f_y = \frac{\partial}{\partial y}\int_{1}^{xy} h(s) ds $$ From the Leibniz Rule, we have: $$f_y = \int_{1}^{xy} \frac{\partial h(s)}{\partial y} ds + h(xy) \frac{\partial (xy)}{\partial y} - h(1) \frac{\partial 1}{\partial y} $$ Again, \(\frac{\partial h(s)}{\partial y}\) is zero because there is no \(y\) in \(h(s)\) and the last term on the right side vanishes: $$f_y = 0 + h(xy) \frac{\partial (xy)}{\partial y} - 0 $$ Now, we just need to find the derivative of \((xy)\) with respect to \(y\): $$\frac{\partial (xy)}{\partial y} = x$$ So, we have: $$f_y = h(xy) x$$
04

Express the Solution

We've found the partial derivatives of \(f(x, y)\) with respect to \(x\) and \(y\). The solution is: $$f_x = h(xy) y$$ $$f_y = h(xy) x$$

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Key Concepts

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

Partial Derivatives
Partial derivatives allow us to examine how a function change with respect to one variable, while keeping all others constant. This is crucial in multi-variable calculus, where functions depend on more than one variable. For instance, consider the function \( f(x, y) = \int_{1}^{xy} h(s) ds \). It depends on both \( x \) and \( y \), so we explore its behavior through partial derivatives.To find \( f_x \), the partial derivative with respect to \( x \), we treat \( y \) as a constant and apply the Leibniz rule, allowing us to simplify and find how \( f(x, y) \) changes as \( x \) varies. Similarly, \( f_y \) is found by considering \( x \) constant and varying \( y \). The results in the exercise demonstrate that:
  • \( f_x = h(xy) y \)
  • \( f_y = h(xy) x \)
Thus, these derivatives tell us about the sensitivity of \( f \) to changes in \( x \) and \( y \).
Continuous Function
A continuous function is one where there are no abrupt jumps or breaks. In simple terms, you can draw it without lifting your pencil from the paper. Continuous functions are essential because they guarantee that integrals, like \( \int_{1}^{xy} h(s) ds \), are well-behaved and meaningful.In the problem, \( h(s) \) is continuous over all real numbers. This means:
  • The values of \( h \, (s) \) change smoothly without sudden jumps.
  • We can explore the integral \( f(x, y) \) confidently, knowing it spans over a smooth curve.
Continuous functions offer the stability needed for calculus operations, making them much easier and reliable to analyze.
Integral Calculus
Integral calculus is all about finding the total accumulation of quantities. When you integrate a function, you're essentially summing up infinitely many infinitesimal parts over a range.In our exercise, we integrated \( h(s) \) over the range from 1 to \( xy \) with respect to \( s \):\[ f(x, y) = \int_{1}^{xy} h(s) \, ds\]This tells us the accumulated effect of \( h(s) \) from 1 to \( xy \).The Leibniz rule helps differentiate such integrals with variable limits, a key tool here. It allows us to calculate how this accumulated value changes as either \( x \) or \( y \) change while respecting the function鈥檚 structure. Integral calculus, supported by concepts like the Leibniz rule, builds the foundation for many physical and technical applications where changes in quantities matter a lot.

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