Problem 59 - Calculus for Biology and Medicine

Question

Differentiate with respect to the independent variable. $$ f(x)=\sqrt{x}(x-1) $$

Step-by-Step Solution

Answer

The derivative of the function is \(\frac{3x - 1}{2\sqrt{x}}\).

Step 1: Identify the Function Type

The function given is a product of two functions: \(f(x) = \sqrt{x} \cdot (x-1)\). This is a product of \(u(x) = \sqrt{x}\) and \(v(x) = (x-1)\). We will use the product rule for differentiation.

Step 2: State the Product Rule

The product rule for differentiation states that if \(y = u(x) \cdot v(x)\), then \(\frac{dy}{dx} = u'(x) \cdot v(x) + u(x) \cdot v'(x)\). Apply this rule to differentiate \(f(x)\).

Step 3: Differentiate \(u(x) = \sqrt{x}\)

Rewrite \(u(x) = x^{1/2}\). The derivative \(u'(x)\) is \(\frac{1}{2}x^{-1/2}\) or \(\frac{1}{2\sqrt{x}}\).

Step 4: Differentiate \(v(x) = (x-1)\)

The function \(v(x) = x - 1\) is a linear function, and its derivative \(v'(x)\) is \(1\).

Step 5: Apply the Product Rule

Using the product rule: \[\frac{df}{dx} = u'(x) \cdot v(x) + u(x) \cdot v'(x)\] \[= \left(\frac{1}{2\sqrt{x}}\right) \cdot (x-1) + \sqrt{x} \cdot 1\].

Step 6: Simplify the Expression

Simplify the expression: \[\frac{df}{dx} = \frac{x-1}{2\sqrt{x}} + \sqrt{x}\]. Combine the terms over a common denominator: \[= \frac{x-1 + 2x}{2\sqrt{x}} = \frac{3x - 1}{2\sqrt{x}}\].

Key Concepts

Product Rule in Calculus

The product rule is a fundamental tool in calculus that helps us differentiate products of two functions. When dealing with differentiation, it's common to encounter functions that can be expressed as the product of two simpler functions. In such cases, directly applying the basic differentiation rules is insufficient. This is where the product rule becomes essential. For functions of the form \(y = u(x) \cdot v(x)\), the product rule provides the formula:

• \( \frac{dy}{dx} = u'(x) \cdot v(x) + u(x) \cdot v'(x) \)

This formula essentially tells us to take the derivative of the first function, multiply it by the second function, and then add it to the first function multiplied by the derivative of the second function. Such a logical and systematic approach ensures that even complex multiplied functions can be differentiated accurately.

Understanding Differentiation

Differentiation is one of the core concepts in calculus and refers to the process of finding the rate at which a function changes at any given point. This process allows us to compute the derivative, which measures the instantaneous rate of change.To perform differentiation correctly, we need to follow established rules and formulas like the power rule, the product rule, and the chain rule. In the case of the given function, applying differentiation manually involves:

• Rewriting functions into differentiable forms, like converting \(\sqrt{x}\) into \(x^{1/2}\).
• Using specific rules that fit the function type, as done with the product rule in our exercise.

The result of successful differentiation yields a new function that predicts how the original function behaves as the inputs shift.

The Importance of Calculus

Calculus plays a pivotal role in various fields, not just in mathematics. It allows us to understand changes and motion, which is crucial for forming predictions and modeling real-world phenomena. Functions that appear complex, such as the provided \(f(x) = \sqrt{x}(x-1)\), can be analyzed for their changing nature through calculus techniques like differentiation.The overarching aim of calculus is to provide a way to quantify change and understand the behavior of functions. It equips us with tools like derivatives and integrals, ensuring that we can solve real-life problems, be it in engineering, physics, economics, or other areas. Techniques such as the product rule serve as building blocks, making challenging problems approachable and solvable.