Problem 60 - Calculus for Biology and Medicine

Question

Differentiate with respect to the independent variable. \(f(x)=\sqrt{x}\left(x^{4}-x^{2}\right)\)

Step-by-Step Solution

Answer

The derivative is \(f'(x) = \frac{9}{2}x^{7/2} - \frac{1}{2}x^{5/2} - 2x^{3/2}\).

Step 1: Rewrite the Function

First, express the function in a form that facilitates differentiation. The function is given as \(f(x) = \sqrt{x}(x^4 - x^2)\). Rewrite \(\sqrt{x}\) as \(x^{1/2}\) to simplify differentiation: \[ f(x) = x^{1/2}(x^4 - x^2) \].

Step 2: Apply the Product Rule

The function is a product of two functions, \(u(x) = x^{1/2}\) and \(v(x) = x^4 - x^2\). The product rule states that the derivative of a product \(uv\) is \(u'v + uv'\). We need to find \(u'\) and \(v'\) to apply this rule.

Step 3: Differentiate Individual Functions

Differentiate \(u(x) = x^{1/2}\):\[ u'(x) = \frac{1}{2}x^{-1/2} = \frac{1}{2\sqrt{x}} \].Differentiate \(v(x) = x^4 - x^2\):\[ v'(x) = 4x^3 - 2x \].

Step 4: Substitute and Simplify

Substitute \(u\), \(v\), \(u'\), and \(v'\) into the product rule formula:\[ f'(x) = \left(\frac{1}{2}x^{-1/2}\right)(x^4 - x^2) + x^{1/2}(4x^3 - 2x) \].Simplify each term:1. \(\frac{1}{2}x^{-1/2}(x^4 - x^2) = \frac{x^{4} - x^2}{2\sqrt{x}} = \frac{x^{7/2} - x^{5/2}}{2} \).2. \(x^{1/2}(4x^3 - 2x) = 4x^{7/2} - 2x^{3/2}\).Thus, \[ f'(x) = \frac{x^{7/2} - x^{5/2}}{2} + 4x^{7/2} - 2x^{3/2} \].

Step 5: Combine and Simplify Like Terms

Combine like terms to further simplify the result:- Combine \(x^{7/2}\) terms: \(\frac{x^{7/2}}{2} + 4x^{7/2} = \frac{1}{2}x^{7/2} + \frac{8}{2}x^{7/2} = \frac{9}{2}x^{7/2}\).- \(x^{5/2}\) and \(x^{3/2}\) terms remain as \(-\frac{1}{2}x^{5/2} - 2x^{3/2}\).The final derivative is \[ f'(x) = \frac{9}{2}x^{7/2} - \frac{1}{2}x^{5/2} - 2x^{3/2} \].

Key Concepts

Product Rule

The product rule is an essential tool in calculus for finding the derivative of a product of two functions. Imagine you have two functions, say \( u(x) \) and \( v(x) \), and you need to differentiate their product, \( u(x) \cdot v(x) \). The product rule helps us by stating the derivative can be found using the formula: \[ \frac{d}{dx}[u(x) \cdot v(x)] = u'(x) \cdot v(x) + u(x) \cdot v'(x) \]

• \( u'(x) \) and \( v'(x) \) represent the derivatives of \( u(x) \) and \( v(x) \), respectively.
• Each term involves multiplying one derivative by the unchanged other function.

This rule is crucial because it simplifies the process of differentiation for products, especially when dealing with more complex functions. Remember, being systematic in how you apply the product rule is key to avoiding mistakes.

Functions of a Variable

When dealing with differentiation, understanding what a function of a variable is can make the process smoother. In mathematical terms, a function is something that takes an input, transforms it, and provides an output. More specifically, a function of a variable, like \( f(x) \), relies on the variable \( x \) to determine its outputs.

• \( f(x) = x^{1/2}(x^4 - x^2) \) is a single function of \( x \) made up of parts that each rely on \( x \).
• The function is filled with terms \( x^4 \), \( x^2 \), and \( x^{1/2} \), expressing how \( x \) is involved differently in each term.

From this perspective, it's simpler to see why individual terms need to be handled separately in differentiation. You deal with one piece at a time, respecting how each depends on \( x \). This approach makes differentiation less daunting because you're breaking it down into bits you can manage.

Derivatives

Derivative is a fundamental concept in calculus and describes how a function changes as its input changes. Think of it as a way of capturing the rate of change or the slope of the function at any given point. For any function \( f(x) \), its derivative, \( f'(x) \), tells us the instantaneous rate of change at a particular \( x \).

• If \( f(x) = x^n \), then the derivative \( f'(x) = nx^{n-1} \) is obtained using the power rule. This tells us how fast \( x^n \) grows or shrinks at any point.
• For more complex functions, apply the product or chain rules as needed.

In the context of the original exercise, finding derivatives meant taking apart the function into smaller segments that could be differentiated separately. Each segment followed standard rules (like the power rule), and all were combined appropriately to form the derivative of the original function. It's a methodology that simplifies handling compound functions, like the one in the exercise, ensuring nothing is left to chance.