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 is \(f'(x) = \frac{3x - 1}{2\sqrt{x}}\).

Step 1: Understand the Function

The function given is a product of two expressions: \(f(x) = \sqrt{x}(x-1)\). We need to differentiate this function using the product rule because it is a product of \(u(x) = \sqrt{x}\) and \(v(x) = (x-1)\).

Step 2: Recall the Product Rule

The product rule states that the derivative of a product of two functions \(u(x)\) and \(v(x)\) is given by: \( (uv)' = u'v + uv' \).

Step 3: Differentiate \(u(x)\)

First, differentiate the function \(u(x) = \sqrt{x}\). Recall that \(\sqrt{x} = x^{1/2}\), and using the power rule, \(u'(x) = \frac{1}{2}x^{-1/2}\).

Step 4: Differentiate \(v(x)\)

Now, differentiate the function \(v(x) = (x-1)\). The derivative of this linear expression is \(v'(x) = 1\).

Step 5: Apply the Product Rule

Use the product rule to find the derivative of the given function: \(f'(x) = u'(x)v(x) + u(x)v'(x)\).Substitute \(u' = \frac{1}{2}x^{-1/2}\), \(v = x - 1\), \(u = \sqrt{x}\), and \(v' = 1\): \[f'(x) = \left(\frac{1}{2}x^{-1/2}\right)(x-1) + \sqrt{x}(1)\].

Step 6: Simplify the Expression

Simplify the expression from the previous step:\[f'(x) = \frac{x-1}{2\sqrt{x}} + \sqrt{x}\].The first term is simplified by noting \(x^{-1/2} = \frac{1}{\sqrt{x}}\) and the second term is already simplified.

Step 7: Combine Terms

To combine the terms, express \(\sqrt{x}\) with a common denominator:\[f'(x) = \frac{x-1}{2\sqrt{x}} + \frac{2x}{2\sqrt{x}}\]. Combine them:\[f'(x) = \frac{x-1 + 2x}{2\sqrt{x}} = \frac{3x - 1}{2\sqrt{x}}\].

Key Concepts

Understanding the Product Rule

When you encounter an expression like \(f(x) = \sqrt{x}(x-1)\), where two functions are multiplied together, the product rule is your go-to tool for differentiation. The product rule is essential when the differentiation of a product of two functions, \(u(x)\) and \(v(x)\), is required. The rule is expressed as:

\((uv)' = u'v + uv'\).

Here’s what this means:

• Differentiate each function separately.
• Multiply the derivative of the first function by the second function.
• Add this to the first function multiplied by the derivative of the second function.

Applying the product rule correctly simplifies our calculations and helps in accurately finding the derivative of composite functions. It is especially useful in higher mathematics and fields such as engineering and physics, where complex functions commonly appear.

Applying the Power Rule for Function Differentiation

To differentiate expressions like \(u(x) = \sqrt{x}\), we use the power rule. The power rule is one of the most straightforward rules for differentiation. It states that if you have a function in the form of \(x^n\), its derivative is \(nx^{n-1}\).

For the function at hand, remember that \(\sqrt{x} = x^{1/2}\). Applying the power rule here, the derivative becomes:

\(u'(x) = \frac{1}{2}x^{-1/2}\).

This rule helps quickly and efficiently in taking derivatives of powers of \(x\), even fractional powers like in this example. It also lays a foundation for understanding more complex derivation techniques, as it captures the fundamental behavior of differentiation with respect to power.

Simplifying the Derivative Expression

Once derivatives of individual components are found using the product and power rules, the next step is to combine and simplify the expressions. In our function:

\[f'(x) = \left(\frac{1}{2}x^{-1/2}\right)(x-1) + \sqrt{x}(1)\],
simplification is crucial for the final derivative to be as clear and usable as possible.

Use the identity \(x^{-1/2} = \frac{1}{\sqrt{x}}\) to reorganize terms.

• For the first component, distribute and simplify: \(\frac{x-1}{2\sqrt{x}}\).
• The second component is already in simplified form: \(\sqrt{x}\).

The final expression: \[f'(x) = \frac{x-1 + 2x}{2\sqrt{x}} = \frac{3x - 1}{2\sqrt{x}}\], represents the simplified version of the derivative, using a common denominator for a neat, clear expression.

Understanding simplification techniques ensures that you present derivatives in their most accessible form, improving clarity and comprehension in mathematical communication.