Problem 3 - Technical Mathematics with Calculus

Question

Derivatives with Respect to Other Variables. If \(w=y^{2}+u^{3},\) find \(d w / d u.\)

Step-by-Step Solution

Answer

\( \frac{dw}{du} = 3u^{2} \)

Step 1: Differentiate with Respect to u

To find \( \frac{dw}{du} \), apply the rules of differentiation to each term of \(w=y^{2}+u^{3}\). The term \(y^{2}\) is treated as a constant because it does not contain the variable \(u\). The term \(u^{3}\) is differentiated normally using the power rule.

Step 2: Apply the Power Rule

The power rule states that the derivative of \(u^{n}\) with respect to \(u\) is \(n\cdot u^{n-1}\). Apply this to \(u^{3}\) to get \(3 \cdot u^{2}\). Since \(y^{2}\) is a constant with respect to \(u\), its derivative is zero.

Step 3: Combine the Results

After differentiating both terms, combine them to get the final derivative of \(w\) with respect to \(u\). The derivative of the constant \(y^{2}\) is zero, and the derivative of \(u^{3}\) is \(3u^{2}\). So, \( \frac{dw}{du} = 0 + 3u^{2} \).

Key Concepts

Differentiation

Differentiation is a fundamental concept in calculus, which allows us to find the rate at which one quantity changes with respect to another. Imagine you're driving a car and your speed is the rate of change of distance as time progresses. Similarly, in mathematics, if we have a function that describes how one variable changes with another—like how the area of a circle changes with its radius—we use differentiation to find the rate of this change at any point.

For the exercise given, when we differentiate the function \(w=y^{2}+u^{3}\) with respect to \(u\), we are essentially looking for how \(w\) changes as \(u\) alone changes, while treating any other variables as constants. This approach is essential in understanding complex relationships between multiple variables in various fields such as physics, economics, and engineering.

Power Rule

The power rule is one of the most basic and widely used rules for differentiation. It is a shortcut that makes finding derivatives easier without needing to resort to the definition of the derivative. According to the power rule, if we have a function \(u^{n}\) where \(n\) is any real number, and we want to find its derivative with respect to \(u\), we simply multiply by the exponent \(n\) and then subtract one from the exponent. The mathematical expression for this rule is: \( \frac{d}{du} u^{n} = n \times u^{n-1} \).

In the context of our exercise, the power rule is applied to the term \(u^{3}\). According to the rule, the derivative of \(u^{3}\) with respect to \(u\) is \(3u^{2}\). This streamlines the calculation, making it easier to handle even more complex equations with higher powers of variables.

Multivariable Calculus

Multivariable calculus extends the concepts of single-variable calculus to functions with more than one variable. It provides tools to analyze mathematical systems that have changeable quantities, each affecting the others. The exercise provided is a simple case of multivariable calculus where the function \(w\) depends on two variables, \(y\) and \(u\).

While taking the derivative of \(w\) with respect to \(u\), we treat \(y\) as a constant because, in this scenario, we are only interested in how \(w\) changes as \(u\) changes. Multivariable calculus can get quite complex as it involves partial derivatives, gradient vectors, and multiple integrals, which are crucial for navigating the terrain of functions with multiple inputs and outputs. Understanding this discipline is vital for fields such as computer graphics, optimization, and physical sciences.