Problem 20 - Calculus: An Integrated Approach to Functions and Their Rates of Change

Question

In Problems 19 through 22, find \(\frac{d y}{d x}\). Take the time to prepare the expression so that it is as simple as possible to differentiate. $$ y=\sqrt{\left(x^{2}+3\right)^{5}} $$

Step-by-Step Solution

Answer

\(\frac{dy}{dx} = 5x(x^{2}+3)^{3/2}\)

Step 1: Rewrite the function

Firstly, the function can be rewritten in a way that allows the differentiation to be more straightforward. The function can be rewritten as \(y=(x^{2}+3)^{5/2}\).

Step 2: Apply Chain Rule

Apply the chain rule for differentiation, which is \(dy/dx = dy/du * du/dx\), with \(u = x^{2}+3\). The derivative of \(y=u^{5/2}\) with respect to \(u\) is \((5/2)u^{3/2}\) and the derivative of \(u=x^{2}+3\) with respect to \(x\) is \(2x\).

Step 3: Compute the derivative

Now, by the chain rule, \(dy/dx = dy/du * du/dx = (5/2)(x^{2}+3)^{3/2} * 2x= 5x(x^{2}+3)^{3/2}\). This is the expression for \(\frac{dy}{dx}\).

Step 4: Simplify the derivative

There is no need of simplification in this particular problem as the derivative is already in the simplest form.

Key Concepts

Chain Rule

The Chain Rule is a crucial concept in calculus used for differentiating composite functions. Imagine a function nested within another, similar to stacking boxes. The Chain Rule helps us differentiate these complex expressions efficiently.

In our exercise, the function is initially given as a square root expression, which is \(y=\sqrt{(x^2+3)^{5}}\). However, to make differentiation easier, we reformulate it into a familiar power form: \((x^2+3)^{5/2}\).

• The outer function is \(u^{5/2}\) - where \(u = x^2+3\).
• The inner function is \(x^2+3\).

To differentiate, the Chain Rule formula is applied: \( \frac{dy}{dx} = \frac{dy}{du} \times \frac{du}{dx} \). This tells us to take the derivative of the outer function concerning the inner function and multiply it by the derivative of the inner function concerning \(x\).

This powerful tool allows us to tackle a wide range of calculus problems that involve compositions of multiple functions, and understanding this rule significantly eases the differentiation process.

Simplifying Expressions

Simplifying expressions is an essential step before carrying out differentiation or any algebraic manipulation. In calculus, functions often come in complex forms, and rewriting them more simply can save a lot of time and effort.

In our exercise with \(y=\sqrt{(x^2+3)^{5}}\), the expression appears complicated. By rewriting it as \((x^2+3)^{5/2}\), we align it with easier differentiation rules pertaining to powers and exponents. This preparation step is vital because:

• It transforms the expression into a form where power rules are straightforward to apply.
• It sets the stage for effectively using the Chain Rule.

Such simplifications often help in understanding the function's behavior better and can lead to clearer solutions. It's a preliminary step that prevents errors and streamlines the whole process of solving calculus problems.

Calculus Problems

Calculus problems require careful breakdown and understanding of functions. Differentiation, especially, is one of the main tools we use to understand how functions change.

Take our problem where we need to find \(\frac{dy}{dx}\) for \(y=\sqrt{(x^2+3)^{5}}\). Calculus helps in dissecting this:

• Identify the expression form and manipulate it for easy differentiation.
• Apply rules systematically, in our case, using the Chain Rule.

Calculus problems often present visually complex structures, but breaking them down into smaller, manageable parts is critical. These problems improve critical thinking and bolster confidence in handling mathematical challenges. As you practice more, you become adept at spotting patterns and applying the most suitable methods to solve them.

Understanding differentiation and these steps is a step into deeper concepts of calculus, which reveals more about the behavior of functions and their practical applications.