Problem 28 - Applied Mathematics: For the Managerial, Life, and Social Sciences

Question

Find the derivative of each function. \(f(u)=(2 u+1)^{3 / 2}+\left(u^{2}-1\right)^{-3 / 2}\)

Step-by-Step Solution

Answer

The derivative of the function with respect to u is: \[f'(u) = 3(2u + 1)^{1/2} - 3u(u^2 - 1)^{-5/2}.\]

Step 1: Rewrite the function and identify the terms

Rewrite the function as \(f(u) = (2u + 1)^{3/2} + (u^2 - 1)^{-3/2}\). There are two terms in this function. The first term is \((2u + 1)^{3/2}\) and the second term is \((u^2 - 1)^{-3/2}\).

Step 2: Find the derivative of the first term using the chain rule

To find the derivative of the first term, we can use the chain rule, which says that the derivative of a composite function is the derivative of the outer function times the derivative of the inner function. The first term can be written as the composite function \(y(u)=(2u+1)\) and \(g(y)=y^{3/2}\). The derivative of \(u\) with respect to \(y\) is simply \(dy/du=3y^{1/2}\). Now, we can apply the chain rule: \[f'(u) = \frac{d}{du} (2u + 1)^{3/2} = \frac{3}{2} (2u + 1)^{1/2} \cdot (2).\]

Step 3: Find the derivative of the second term using the chain rule

Similarly, to find the derivative of the second term, we can use the chain rule again. The second term can be written as the composite function \(y(u)=(u^2-1)\) and \(g(y)=y^{-3/2}\). The derivative of \(y\) with respect to \(u\) is\(dy/du=2u\). Now, we can apply the chain rule: \[f'(u) = \frac{d}{du} (u^2 - 1)^{-3/2} = -\frac{3}{2}(u^2 - 1)^{-5/2} \cdot (2u).\]

Step 4: Combine the derivatives of both terms

Finally, we can add the derivatives of both terms to find the derivative of the whole function: \[f'(u) = \frac{3}{2} (2u + 1)^{1/2} \cdot (2) - \frac{3}{2}(u^2 - 1)^{-5/2} \cdot (2u).\] The derivative of the function with respect to u is: \[f'(u) = 3(2u + 1)^{1/2} - 3u(u^2 - 1)^{-5/2}.\]

Key Concepts

Understanding the Chain Rule

The chain rule is an essential technique in calculus for finding the derivative of composite functions. It helps to differentiate a function by breaking it down into its inner and outer functions. Simply put, if you have a function inside another function, the chain rule is your best tool.

To apply the chain rule, you first differentiate the outer function and then multiply it by the derivative of the inner function. Imagine having a function like \(g(f(x))\), where \(g\) is the outer function and \(f(x)\) is the inner one. According to the chain rule, the derivative of \(g(f(x))\) is \[g'(f(x)) \cdot f'(x)\].

Essentially, you follow these steps:

• Identify the inner function and the outer function.
• Differentiate the outer function.
• Differentiate the inner function.
• Multiply these derivatives together.

The beauty of the chain rule lies in its ability to simplify complex problems by "chaining" processes together. Makes sense, right? Let's see how it works with composite functions.

Composite Functions in Calculus

A composite function is a function that results from combining two or more functions. When you see expressions like \(f(u) = (2u + 1)^{3/2} + (u^2 - 1)^{-3/2}\), it means you're dealing with composite functions. Think of them as layers you peel back to find the derivative using rules like the chain rule.

In our case, consider these two separate parts:

• The first term \( (2u + 1)^{3/2} \) consists of the inner function \(2u + 1\) inside the outer function \(y^{3/2}\).
• The second term \( (u^2 - 1)^{-3/2} \) consists of the inner function \(u^2 - 1\) inside \(y^{-3/2}\).

Composite functions require more finesse during differentiation due to their layered structure.

The strategy involves unraveling the layers and applying the appropriate rules for each part. This process highlights the application of the chain rule since many composite functions often need it to take derivatives effectively.

Differentiation Techniques

Differentiation techniques are strategies used in calculus to determine the derivative of functions, which indicate how the function changes at any given point. Trying to differentiate straightforward functions is often easier, but when encountering functions composed of multiple parts or exhibiting complex behavior, advanced techniques like the chain rule are invaluable.

In tackling derivatives, here's what you can typically rely on:

• Power Rule: For functions like \(x^n\), the derivative is \(nx^{n-1}\).
• Product Rule: Used when differentiating a product of two functions.
• Quotient Rule: Useful for the derivative of a quotient of two functions.
• Chain Rule: Perfect for composite functions, as we've discussed.

In our derivative example, the chain rule is at the forefront, simplifying each term in the composite function. This technique turns complicated terms into manageable pieces, making the differentiation process systematic and organized.

As you explore a variety of functions, having a solid understanding of these methods allows you to choose the best approach effectively and accurately. With practice, determining how functions transform can become intuitive even with complex structures.