Problem 28 - Thomas Calculus

Question

Find the derivatives of the functions in Exercises \(19-38\) $$ y=(5-2 x)^{-3}+\frac{1}{8}\left(\frac{2}{x}+1\right)^{4} $$

Step-by-Step Solution

Answer

\( \frac{dy}{dx} = 6(5-2x)^{-4} - \frac{1}{x^2}\left(\frac{2}{x} + 1\right)^3 \)

Step 1: Differentiate the First Term

The first part of the function is \((5-2x)^{-3}\). To differentiate it, apply the power rule and chain rule. The derivative of \(u^n\) is \(n \cdot u^{n-1} \cdot \frac{du}{dx}\), where \(u = 5-2x\) and \(n = -3\). This gives: \[ \frac{d}{dx}(5-2x)^{-3} = -3(5-2x)^{-4} \cdot (-2) = 6(5-2x)^{-4} \]

Step 2: Differentiate the Second Term

The second part of the function is \(\frac{1}{8}\left(\frac{2}{x}+1\right)^{4}\). Again, apply the power rule and chain rule. Here, \(v = \frac{2}{x} + 1\) and the derivative of \(v^m\) (where \(m = 4\)) gives: \[ \frac{d}{dx}\left( \left(\frac{2}{x} + 1\right)^4 \right) = 4\left( \frac{2}{x} + 1 \right)^3 \cdot \left( \frac{d}{dx}\left(\frac{2}{x}\right) \right) \]For \(\frac{d}{dx}\left(\frac{2}{x}\right)\), use the rule that \(\frac{d}{dx}(x^{-1}) = -x^{-2}\), meaning that \(\frac{d}{dx}\left(\frac{2}{x}\right) = -2x^{-2}\), giving us:\[ \frac{d}{dx}\left( \frac{2}{x} + 1 \right) = -\frac{2}{x^2} \]Thus, the derivative of the second term is:\[ \frac{1}{8} \cdot 4 \left( \frac{2}{x} + 1 \right)^3 \cdot \left(-\frac{2}{x^2}\right) = -\frac{1}{x^2} \left( \frac{2}{x} + 1 \right)^3 \]

Step 3: Combine the Derivatives

Now, combine the derivatives of the two terms to find the overall derivative of the function:\[ \frac{dy}{dx} = 6(5-2x)^{-4} - \frac{1}{x^2}\left(\frac{2}{x} + 1\right)^3 \]

Key Concepts

Chain Rule

The chain rule is a powerful tool in calculus that allows us to find the derivative of composite functions. Composite functions are those that are formed by combining two or more functions. In simpler terms, the chain rule helps when one function is inside another function. This is common in many calculus problems.

The rule itself can be described as:

• If you have a function that is the composition of two functions, say \( f(g(x)) \), the derivative is \( f'(g(x)) \cdot g'(x) \).
• This means you first differentiate the outer function while keeping the inner function the same and multiply it by the derivative of the inner function.

Understanding the chain rule is crucial for tackling more complex differentiation problems. It’s the principle that was used to differentiate both terms of the given function in the problem.

Power Rule

The power rule is one of the straightforward rules in differentiation, commonly used when the function is a power of x. It’s easy to remember and apply. The power rule states:

• If you have a function in the form of \( x^n \), the derivative is \( n \cdot x^{n-1} \).

In the exercise, the power rule was applied in conjunction with the chain rule. For example, in the term \((5-2x)^{-3}\), the power rule helped determine how to handle the \(-3\) exponent by "bringing it down" and reducing the power by one. The same process occurs with the \( \left(\frac{2}{x} + 1\right)^4 \) term.

Use the power rule whenever you see a function raised to a power, it makes the process of finding derivatives swift.

Differentiation

Differentiation is a process in calculus, used to find the rate at which one quantity changes with respect to another. It's a fundamental concept central to calculus, and helps in understanding the behavior of functions.

When working through differentiation problems, you often will apply several rules such as the product rule, quotient rule, chain rule, and power rule depending on the structure of the problem. In this particular exercise, the main focus was:

• Applying the chain rule and power rule to separate parts of a compound function.
• Carefully combining derivatives step-by-step to ensure no elements are missed out.

Differentiation is used widely across many fields for analyzing trends, data modeling, physics, and any field that involves change. Mastering differentiation, and the rules that come with it, is essential for success in calculus.