Problem 42 - Calculus for Biology and Medicine

Question

In Problems \(39-48\), find the indicated partial derivatives. $$ f(x, y)=\sin (x-y) ; \frac{\partial^{2} f}{\partial y \partial x} $$

Step-by-Step Solution

Answer

The mixed partial derivative \( \frac{\partial^2 f}{\partial y \partial x} \) is \( \sin(x-y) \).

Step 1: Find the first partial derivative with respect to x

We start by finding the first partial derivative of the function \( f(x, y) = \sin(x-y) \) with respect to \( x \).\[ \frac{\partial f}{\partial x} = \cos(x-y) \cdot \frac{\partial}{\partial x}(x-y) = \cos(x-y)\cdot 1 = \cos(x-y) \]

Step 2: Differentiate with respect to y

Now, find the partial derivative of \( \frac{\partial f}{\partial x} = \cos(x-y) \) with respect to \( y \).\[ \frac{\partial}{\partial y} (\cos(x-y)) = -\sin(x-y)\cdot \frac{\partial}{\partial y}(x-y) = -\sin(x-y)(-1) = \sin(x-y) \] This gives us the second mixed partial derivative \( \frac{\partial^2 f}{\partial y \partial x} = \sin(x-y) \).

Key Concepts

Mixed Partial Derivatives

Mixed partial derivatives are crucial in mathematics, especially when dealing with functions with multiple variables. Essentially, they are partial derivatives taken in a sequence, often in differing order.
In the problem given, you first find the derivative of a function with respect to one variable (here, "x"), and then you differentiate the result with respect to another ("y").
An important property of mixed partial derivatives is that if the function and derivatives are continuous and well-defined, the order of differentiation does not matter. This is known as Clairaut's theorem. So, \( \frac{\partial^2 f}{\partial y \partial x} \) is the same as \( \frac{\partial^2 f}{\partial x \partial y} \).
This concept is especially important in physics and engineering where they describe how a system changes in multiple dimensions or directions at once.

Trigonometric Functions

Trigonometric functions, like sine and cosine, play a prominent role in calculus problems, including differentiation. These functions arise naturally in various contexts, such as oscillations and waves.
In our specific problem, the function \( f(x, y) = \sin(x-y) \) involves the "sine" function, which depends on the difference of two variables, \( x \) and \( y \).
When finding partial derivatives of trigonometric functions, remember:

• The derivative of \( \sin(u) \) with respect to \( u \) is \( \cos(u) \).
• The derivative of \( \cos(u) \) is \(-\sin(u) \).

These rules allow us to systematically compute derivatives as shown in the solution by applying the chain rule effectively.

Calculus Problem Solving

Calculus problem solving generally involves a methodical approach: breaking down complex problems into simpler steps. Let's focus on executing this process:

• **Understand the problem:** Clearly define what is being asked, in this case, the mixed partial derivative.
• **Identify the function and variables:** Take into account what functions need to be differentiated and with respect to which variables.
• **Apply differentiation rules:** Use correct rules of differentiation for each function, here using trigonometric differentiation rules.
• **Answer verification:** Ensure that intermediate results are consistent by verifying calculations at each step.

By following these steps, tackling even seemingly complex problems, like those involving mixed partials, becomes manageable. Incorporating a stepwise problem-solving methodology enhances accuracy and allows for manageable, replicable steps across various problems.