Problem 5 - Calculus for Biology and Medicine

Question

In Problems 1-16, find \(\partial f / \partial x\) and \(\partial f / \partial y\) for the given functions. $$ f(x, y)=\sin (x+y) $$

Step-by-Step Solution

Answer

\( \frac{\partial f}{\partial x} = \cos(x+y) \) and \( \frac{\partial f}{\partial y} = \cos(x+y) \).

Step 1: Understanding the Problem

We need to find the partial derivatives of the function \( f(x, y) = \sin(x + y) \) with respect to \( x \) and \( y \). This involves treating all other variables constant except for the one we are differentiating with respect to.

Step 1: Differentiate with Respect to x

To find \( \frac{\partial f}{\partial x} \), treat \( y \) as a constant and differentiate \( \sin(x+y) \) with respect to \( x \) using the chain rule. The derivative of \( \sin(u) \) with respect to \( u \) is \( \cos(u) \), so \( \frac{\partial}{\partial x}\sin(x+y) = \cos(x+y) \cdot \frac{\partial}{\partial x}(x+y) = \cos(x+y) \cdot (1) = \cos(x+y) \).

Step 2: Differentiate with Respect to y

To find \( \frac{\partial f}{\partial y} \), treat \( x \) as a constant and differentiate \( \sin(x+y) \) with respect to \( y \). Again, applying the chain rule, \( \frac{\partial}{\partial y} \sin(x+y) = \cos(x+y) \cdot \frac{\partial}{\partial y}(x+y) = \cos(x+y) \cdot (1) = \cos(x+y) \).

Key Concepts

Multivariable Calculus

Multivariable calculus is a branch of mathematics that extends calculus concepts to functions of multiple variables. Unlike single-variable calculus, where functions depend on one variable, multivariable calculus deals with functions like \[ f(x, y) = \sin(x + y) \]where the output depends on two or more input variables, in this case, \( x \) and \( y \). When working with such functions, it's important to consider how changes in each input variable affect the function's output.

For functions of several variables, the concept of partial derivatives is essential. These derivatives measure how the function changes as one of the input variables changes, while the others are kept constant. This is akin to finding a slope in single-variable calculus but now done in a multiple-parameter space.

• When differentiating with respect to one variable, consider all other variables as constants.
• This allows us to explore the rate of change in one direction of the multi-dimensional space.

Understanding partial derivatives leads to applications in optimization, economics, physics, and engineering, where multiple factors influence outcomes simultaneously. Knowing how to handle each variable independently in a function helps in predicting changes and making informed decisions.

Chain Rule

The chain rule is a crucial tool in calculus, especially when dealing with compositions of functions. In the context of multivariable calculus, it becomes essential for finding partial derivatives of composite functions. For our function, \[ f(x, y) = \sin(x + y) \]applying the chain rule lets us differentiate the sine function, which itself is "wrapped" around the linear function \( x + y \).

The chain rule states that if a function \( f \) depends on an intermediate variable \( u \), which in turn depends on \( x \), then:\[ \frac{df}{dx} = \frac{df}{du} \cdot \frac{du}{dx} \]Using the chain rule in partial differentiation, we consider each path through which one variable affects the function. For calculating \( \frac{\partial f}{\partial x} \), we treat \( y \) as constant:

• The derivative of \( \sin(u) \) with respect to \( u \) is \( \cos(u) \).
• The derivative of \( x + y \) with respect to \( x \) is 1.

Thus, using the chain rule, we get:\[ \frac{\partial}{\partial x}\sin(x+y) = \cos(x+y) \cdot 1 = \cos(x+y) \]

The same process is applied for \( \frac{\partial f}{\partial y} \) as well, showing the utility of the chain rule in simplifying these expressions.

Trigonometric Functions

Trigonometric functions, like sine and cosine, play a significant role in calculus due to their periodic properties and derivatives. Here, the function \[ \sin(x + y) \]is a trigonometric function that is used widely in various branches of science and engineering.

The derivatives of trigonometric functions are fundamental in solving calculus problems:

• Derivative of \( \sin(u) \) is \( \cos(u) \).
• The periodic nature of these functions helps in modeling wave behaviors and cycles in physics.

For the partial derivatives:- When we differentiate \( \sin(x + y) \) with respect to \( x \), we treat \( y \) as a constant, resulting in \( \cos(x + y) \).
- Likewise, differentiating with respect to \( y \), treating \( x \) as a constant, results again in \( \cos(x + y) \).

These properties showcase how understanding the basic derivatives of sine and cosine can simplify and solve complicated multivariable calculus problems, by recognizing patterns and utilizing the periodic nature of trigonometric functions.