Problem 13 - Calculus for Biology and Medicine

Question

Differentiate the functions with respect to the independent variable. \(f(x)=\frac{\sqrt{2 x-1}}{(x-1)^{2}}\)

Step-by-Step Solution

Answer

The derivative is \(\frac{-3}{(x-1)^3 \cdot \sqrt{2x-1}}\).

Step 1: Recognize the Need for the Quotient Rule

The function you need to differentiate is given in the form of a fraction: \[f(x) = \frac{\sqrt{2x-1}}{(x-1)^{2}}\]In calculus, when you have a function that is a ratio of two functions, the Quotient Rule is typically used.

Step 2: Apply the Quotient Rule

The Quotient Rule states: \[(f/g)' = \frac{g \cdot f' - f \cdot g'}{g^2}\]Identify \(f(x) = \sqrt{2x - 1}\) and \(g(x) = (x-1)^2\). Differentiate both functions separately.

Step 3: Differentiate the Numerator

The numerator \(f(x) = \sqrt{2x - 1}\) can be rewritten as \((2x - 1)^{1/2}\). Using the chain rule:\[f'(x) = \frac{1}{2} (2x - 1)^{-1/2} \cdot 2 = \frac{1}{\sqrt{2x-1}}\]

Step 4: Differentiate the Denominator

The denominator is \(g(x) = (x-1)^2\). Apply the power rule to differentiate:\[g'(x) = 2(x-1) \cdot 1 = 2(x-1)\]

Step 5: Substitute into the Quotient Rule

Insert the differentiated functions into the Quotient Rule:\[\frac{((x-1)^2) \cdot \left(\frac{1}{\sqrt{2x-1}}\right) - (\sqrt{2x-1}) \cdot (2(x-1))}{((x-1)^2)^2}\]

Step 6: Simplify the Expression

To simplify:- The numerator becomes:\[(x-1)^2 \cdot \frac{1}{\sqrt{2x-1}} - 2(x-1)\sqrt{2x-1}\]- The denominator is:\[(x-1)^4\]Further combine and simplify the terms to get \(\frac{(x-1) \cdot [(x-1)-2(2x-1)]}{(x-1)^4 \cdot \sqrt{2x-1}} \).

Step 7: Final Expression

Simplifying further, you obtain:\[\frac{-(3x-3)}{(x-1)^3 \cdot \sqrt{2x-1}}\]Thus, the derivative of the given function \(f(x)\) is:\[\frac{-3}{(x-1)^3 \cdot \sqrt{2x-1}}\] after simplifying.

Key Concepts

Quotient Rule

When you work with a fraction of two functions and want to find its derivative, the Quotient Rule in calculus is your go-to tool. It specifies how to differentiate ratios efficiently. Given a function \(f(x) = \frac{u(x)}{v(x)}\), the rule is mathematically expressed as:

• \((f/g)' = \frac{v(x) \cdot u'(x) - u(x) \cdot v'(x)}{[v(x)]^2}\)

This means you take the derivative of the numerator and multiply it by the denominator, then subtract the numerator multiplied by the derivative of the denominator.
This result is then divided by the denominator squared. It efficiently handles fractions without unnecessary mistakes.
For example, if \(u(x)\) is \(\sqrt{2x-1}\) and \(v(x)\) is \((x-1)^2\), applying the Quotient Rule provides a systematic approach to handling complex fraction derivatives.
Let's explore other differentiation rules used as part of the process by looking into the Chain Rule next.

Chain Rule

The Chain Rule is essential when dealing with compositions of functions. It allows you to differentiate a function that is nested within another function.
Think of it as a "function inside a function." The rule is expressed as:

• If \(y = f(g(x))\), then \(\frac{dy}{dx} = f'(g(x)) \cdot g'(x)\)

In simpler terms, you differentiate the external function while keeping the internal one intact, then multiply by the derivative of the internal function.
For instance, consider \(\sqrt{2x-1}\) rewritten as \((2x-1)^{1/2}\). Here, the inside function \(g(x)\) is \(2x-1\) and the outer function \(f(u) = u^{1/2}\).
Using the Chain Rule, you first derive \((2x-1)^{1/2}\) as \( \frac{1}{2}(2x-1)^{-1/2} \), then multiply by the derivative of \(2x-1\), which is 2.
This results in the derivative being \(\frac{1}{\sqrt{2x-1}}\). This method streamlines differentiation of nested functions.

Power Rule

The Power Rule is one of the most straightforward rules in calculus. It offers a quick way to differentiate functions where a variable is raised to a power. The general formula for the Power Rule is:

• If \(y = x^n\), then \(\frac{dy}{dx} = n \cdot x^{n-1}\)

Simply put, bring down the exponent as a coefficient and reduce the power by one.
In the example, the term \((x-1)^2\) was differentiated using the Power Rule. Here, the power \(n\) is 2.
The differentiation becomes \(2(x-1)^{1}\), not forgetting to apply the Chain Rule for the inner derivative \((x-1)' = 1\).
The powerful combination of these rules turns a potentially tedious process into a more efficient one, enabling quick calculation of derivatives. By exploring these rules, you'll handle any calculus differentiation problem with confidence and clarity.