Question

Find all first partial derivatives of each function. \(f(x, y)=(2 x-y)^{4}\)

Short answer

The partial derivatives are \(8(2x-y)^3\) for \(x\) and \(-4(2x-y)^3\) for \(y\).

Step-by-step solution

Identify the Function

The given function is \(f(x, y) = (2x - y)^{4}\). We need to find its partial derivatives with respect to \(x\) and \(y\).

Calculate the Partial Derivative with respect to x

To find \(\frac{\partial f}{\partial x}\), treat \(y\) as a constant. Use the chain rule. Let \(u = 2x - y\), thus \(f = u^4\). The derivative of \(u^4\) with respect to \(u\) is \(4u^3\), and the derivative of \(u = 2x - y\) with respect to \(x\) is 2. Therefore, \(\frac{\partial f}{\partial x} = 4(2x-y)^3 \cdot 2 = 8(2x-y)^3\).

Calculate the Partial Derivative with respect to y

To find \(\frac{\partial f}{\partial y}\), treat \(x\) as a constant. Again, use the chain rule. With \(u = 2x - y\), \(f = u^4\). The derivative of \(u^4\) with respect to \(u\) is \(4u^3\), and the derivative of \(u = 2x - y\) with respect to \(y\) is -1. Therefore, \(\frac{\partial f}{\partial y} = 4(2x-y)^3 \cdot (-1) = -4(2x-y)^3\).

Key concepts

Chain Rule

The chain rule is a fundamental technique in calculus used to differentiate composite functions. When you're dealing with complex functions, especially those that are composed of other functions, the chain rule becomes invaluable.
It allows us to break down the differentiation process into manageable steps. For instance, if you have a function like \(f(g(x))\), where you need to find the derivative, the chain rule guides you. It states that you should calculate the derivative of the outer function \(f\) with respect to the inner function \(g\), and then multiply by the derivative of the inner function \(g\) with respect to \(x\). In mathematical terms, this is expressed as:

• \(\frac{d}{dx}[f(g(x))] = f'(g(x))\cdot g'(x))\)

In the context of partial derivatives, the chain rule functions similarly. It helps find derivatives by identifying a substitution for expression simplification before applying the rule.

Multivariable Calculus

Multivariable calculus extends the principles of calculus to functions of multiple variables, usually two or more variables. Unlike functions of a single variable, these functions take points from a plane (or higher-dimensional space). This allows us to explore and solve problems in more complex systems.
In functions of several variables, such as \(f(x, y)\), we are interested in how the function changes as each variable changes independently of the others. This leads us to the concept of partial derivatives, where each partial derivative corresponds to the effect of changing one variable while keeping the others constant.
Understanding multivariable calculus is crucial in many scientific fields, including physics and engineering, where systems are often influenced by more than one factor. For example, the temperature of an engine, influenced by both pressure and humidity, can be modeled and analyzed using multivariable calculus.

Derivative with Respect to x

The partial derivative with respect to \(x\) of a function \(f(x, y)\) involves holding the other variable, \(y\), constant while differentiating.
This process reveals how the function \(f\) changes as \(x\) changes, which is key in fields like optimization and design. To calculate it, you treat \(y\) as a constant and find the derivative using standard differentiation rules.
In the given function \(f(x, y) = (2x-y)^4\), we found the partial derivative with respect to \(x\) as follows:

• Identify \(u = 2x - y\)
• Find \(\frac{d}{dx}[(2x - y)^4] = 8(2x-y)^3\)

This tells us that for changes in \(x\), the function changes according to the degree and coefficient found through this differentiation process.

Derivative with Respect to y

Similar to deriving with respect to \(x\), the partial derivative with respect to \(y\) is about understanding how the function changes as \(y\) varies, while \(x\) remains constant.
This type of derivative is essential in creating models that accurately predict the influence of one variable while isolating others.
For the function \(f(x, y) = (2x-y)^4\), we calculate the partial derivative with respect to \(y\) like this:

• Designate \(u = 2x - y\)
• Find \(\frac{d}{dy}[(2x - y)^4] = -4(2x-y)^3\)

This result shows that alterations in \(y\) greatly influence the function, and it also emphasizes the direction indicated by the negative sign, providing insights into the function's behavior when \(y\) alone is varied.