Problem 49 - Applied Mathematics: For the Managerial, Life, and Social Sciences

Question

In Exercises 49-54, find \(\frac{d y}{d u^{\prime}} \frac{d u}{d x^{\prime}}\) and \(\frac{d y}{d x}\). \(y=u^{4 / 3}\) and \(u=3 x^{2}-1\)

Step-by-Step Solution

Answer

The short version of the answer is: \(\frac{dy}{dx} =\frac{4}{3} (3x^2 - 1)^{\frac{1}{3}} * 6x\)

Step 1: Find the derivative of y with respect to u (dy/du')

To find the derivative dy/du', we have the given function: \(y = u^{4/3}\) Now differentiate y with respect to u: \(\frac{dy}{du'} = \frac{4}{3} u^{\frac{4}{3} - 1}\)

Step 2: Find the derivative of u with respect to x (du/dx')

To find the derivative du/dx', we have the given function: \(u = 3x^2 - 1\) Now differentiate u with respect to x: \(\frac{du}{dx'} = 6x\)

Step 3: Apply the chain rule to find the derivative of y with respect to x (dy/dx)

We will now use the chain rule formula (dy/dx) = (dy/du') * (du/dx') to find the derivative of y with respect to x. Substitute the values from step 1 and step 2: \(\frac{dy}{dx} = (\frac{4}{3} u^{\frac{1}{3}}) * (6x)\) Now, substitute \(u = 3x^2 - 1\) back into the equation: \(\frac{dy}{dx} = \frac{4}{3} (3x^2 - 1)^{\frac{1}{3}} * 6x\) So the derivative of y with respect to x is: \(\frac{dy}{dx} =\frac{4}{3} (3x^2 - 1)^{\frac{1}{3}} * 6x\)

Key Concepts

Derivative Calculation

A derivative represents how a function changes as its input changes. It is a core concept in calculus, primarily used to find the rate at which one quantity changes with respect to another. In this exercise, we apply derivative calculation to find how the function \(y = u^{4/3}\) changes with respect to \(u\), and similarly, how \(u = 3x^2 - 1\) changes with respect to \(x\).

• When differentiating \(y = u^{4/3}\) with respect to \(u\), we apply the power rule, which says \(\frac{d}{du}[u^n] = nu^{n-1}\). Here, \(n\) is \(4/3\). Therefore, \(\frac{dy}{du} = \frac{4}{3} u^{1/3}\).
• Similarly, when differentiating \(u = 3x^2 - 1\) with respect to \(x\), we apply the basic derivative rules for polynomials. The derivative of \(3x^2\) is \(6x\), and the derivative of a constant is zero, so \(\frac{du}{dx} = 6x\).

These calculations are foundational for exploring how changes in one variable affect another in a given function. As we proceed, we can use these derivatives to solve more complex problems.

Mathematical Problem Solving

Mathematical problem-solving involves using logical reasoning and mathematical techniques to find solutions to given problems. In this exercise, we use the chain rule, a fundamental principle in calculus, to find a composite derivative. The chain rule helps us differentiate a function that is composed of other functions. Here, \(y\) is a function of \(u\), and \(u\) is a function of \(x\). The chain rule formula is: \[ \frac{dy}{dx} = \frac{dy}{du} \times \frac{du}{dx} \]

• First, find \(\frac{dy}{du}\) and \(\frac{du}{dx}\) independently as described in the derivative calculation section.
• Next, multiply these derivatives together using the chain rule: \(\frac{dy}{dx} = \frac{4}{3}u^{1/3} \times 6x\).
• Replace \(u\) with \(3x^2 - 1\) to express the derivative solely in terms of \(x\): \(\frac{dy}{dx} = \frac{4}{3}(3x^2-1)^{1/3} \times 6x\).

This problem highlights how foundational knowledge of calculus principles aids in solving more intricate mathematical challenges.

Calculus Exercises

Calculus exercises play a crucial role in understanding and mastering concepts like derivatives and the chain rule. These exercises challenge your comprehension and ability to apply the correct rules to find solutions. Such exercises help reinforce how different mathematical rules and techniques interconnect. By working through problems, you develop:

• A better grasp of how derivatives are used to describe the behavior and change of functions.
• An understanding of how to manipulate and transform equations using calculus principles.

In this specific exercise, substituting back variables, like \(u = 3x^2 - 1\), not only enriches your understanding of function compositions but also strengthens your ability to see solutions holistically. Practice with similar calculus problems will improve your problem-solving skills and provide a deeper comprehension of these mathematical concepts.