Problem 20 - Calculus and its applications

Question

Find \(f_{x}\) and \(f_{y}\). $$f(x, y)=3(2 x+y-5)^{2}$$

Step-by-Step Solution

Answer

\( f_x = 12(2x+y-5) \) and \( f_y = 6(2x+y-5) \).

Step 1: Differentiate with respect to x

We start by finding the partial derivative of the function \(f(x, y) = 3(2x+y-5)^2\) with respect to \(x\). We use the chain rule for differentiation. Let \(u = 2x+y-5\), so \(f(x, y) = 3u^2\). The derivative of \(3u^2\) with respect to \(u\) is \(6u\). The derivative of \(u\) with respect to \(x\) is \(2\). Therefore, \( \frac{\partial f}{\partial x} = 6u \cdot 2 = 12(2x+y-5) \).

Step 2: Differentiate with respect to y

Now, we find the partial derivative of the function \(f(x, y) = 3(2x+y-5)^2\) with respect to \(y\). Again, using the chain rule, let \(u = 2x+y-5\), so \(f(x, y) = 3u^2\). The derivative of \(3u^2\) with respect to \(u\) is \(6u\). The derivative of \(u\) with respect to \(y\) is \(1\). Therefore, \( \frac{\partial f}{\partial y} = 6u \cdot 1 = 6(2x+y-5) \).

Key Concepts

Chain Rule

The Chain Rule is a fundamental technique in calculus used to differentiate composite functions. When a function is composed of other functions, applying the Chain Rule helps in finding the derivative with respect to a specific variable. In our exercise, the function is expressed as \( f(x, y) = 3(2x+y-5)^2 \), which is a composite of the outer function \( 3u^2 \) and the inner function \( u = 2x + y - 5 \).

• To differentiate the outer function, we first need the derivative of \( 3u^2 \) with respect to \( u \). This derivative is \( 6u \).
• Next, we find the derivative of the inner function with respect to the variable we are interested in (either \( x \) or \( y \)).
• For \( x \): The derivative of \( u \) with respect to \( x \) is 2, making the partial derivative \( \frac{\partial f}{\partial x} = 6u \cdot 2 \).
• For \( y \): The derivative of \( u \) with respect to \( y \) is 1, giving us \( \frac{\partial f}{\partial y} = 6u \cdot 1 \).

This two-step approach makes it easier to handle functions where direct differentiation isn't straightforward.

Differentiation

Differentiation is the process of finding a derivative, which quantifies how a function changes as its input changes. In multivariable calculus, we often deal with functions of more than one variable.
Differentiating such functions involves finding partial derivatives. These partial derivatives illustrate how the function changes with respect to one variable while keeping the other variables constant.
In the exercise scenario, we are looking for partial derivatives \( f_x \) and \( f_y \):

• \( f_x \) tells us how the function \( f(x, y) \) changes as \( x \) changes, holding \( y \) constant.
• \( f_y \) shows the rate of change of \( f(x, y) \) with respect to \( y \), while keeping \( x \) constant.

The partials are obtained using the Chain Rule, which effectively reduces the complexity by breaking down the multi-layered function into manageable parts.

Multivariable Calculus

Multivariable Calculus extends the principles of single-variable calculus to functions of more than one variable. This area of mathematics allows us to explore how changes in variables impact multivariable functions.
In our example, \( f(x, y) = 3(2x+y-5)^2 \), involves two variables, \( x \) and \( y \), which interact through the function.

• The derivatives \( f_x \) and \( f_y \) help us understand the behavior of \( f \) in various directions of the \( xy \)-plane.
• Partial derivatives are crucial concepts, providing insights into the slope of the function in different directions.
• Applications abound in physics, engineering, computer graphics, and economics, where real-world problems often hinge on understanding multi-variable relationships.

This transformative approach broadens the scope of calculus, making it a powerful tool in analyzing complex systems where different variables come into play.