Problem 13 - Calculus for Biology and Medicine

Question

In Problems 1-28, differentiate the functions with respect to the independent variable. $$ f(x)=\frac{\sqrt{2 x-1}}{(x-1)^{2}} $$

Step-by-Step Solution

Answer

Differentiate using quotient and chain rules to get \( f'(x) = \frac{(x-1)^2 - 2(x-1)\sqrt{2x-1}}{(x-1)^4\sqrt{2x-1}} \).

Step 1: Identify the Differentiation Rule Needed

The function given is a quotient, therefore we need to use the quotient rule, which is: if you have a function \( f(x) = \frac{u(x)}{v(x)} \), then the derivative \( f'(x) \) is given by \( \frac{u'(x)v(x) - u(x)v'(x)}{[v(x)]^2} \).

Step 2: Define the Numerator and Denominator Functions

Identify the numerator function \( u(x) = \sqrt{2x-1} \) and the denominator function \( v(x) = (x-1)^2 \).

Step 3: Differentiate the Numerator Function

To differentiate \( u(x) = \sqrt{2x-1} \), rewrite it as \( u(x) = (2x-1)^{1/2} \) and use the chain rule. The derivative is: \( u'(x) = \frac{1}{2}(2x-1)^{-1/2} \cdot 2 = \frac{1}{\sqrt{2x-1}} \).

Step 4: Differentiate the Denominator Function

Differentiate \( v(x) = (x-1)^2 \) using the power rule: \( v'(x) = 2(x-1)^1 \).

Step 5: Apply the Quotient Rule

Plug \( u(x), u'(x), v(x), \) and \( v'(x) \) into the quotient rule.\[ f'(x) = \frac{\frac{1}{\sqrt{2x-1}} \cdot (x-1)^2 - \sqrt{2x-1} \cdot 2(x-1)}{[(x-1)^2]^2} \]

Step 6: Simplify the Derivative Expression

Simplify the expression:\[ f'(x) = \frac{(x-1)^2 - 2(x-1)\sqrt{2x-1}}{(x-1)^4\sqrt{2x-1}} \].

Key Concepts

Quotient Rule

When you encounter a function that is the division of two other functions, the quotient rule is the tool you'll use to differentiate it. The rule is structured to handle these more complex functions by taking into account both the numerator and the denominator and their derivatives.
This is how the quotient rule works in practice:

• If you have a function defined as \( f(x) = \frac{u(x)}{v(x)} \), the derivative \( f'(x) \) is \( \frac{u'(x)v(x) - u(x)v'(x)}{(v(x))^2} \).
• \( u'(x) \) and \( v'(x) \) are the derivatives of the numerator and denominator functions, \( u(x) \) and \( v(x) \), respectively.
• The quotient rule is particularly useful because it automatically accounts for the product and chain rules when necessary within its formulation.
• The denominator \((v(x))^2\) ensures we factor in the effect of division more extensively, as divisions can introduce faster increases or decreases in the function value.

Remember, while the quotient rule can look daunting due to multiple operations involved, breaking it down into steps as done here makes the process much smoother.

Chain Rule

The chain rule is essential when dealing with composite functions - that is, functions within functions. Imagine peeling layers of an onion, the chain rule allows us to differentiate through these layers effectively.

How does it work? Let's dive in:

• If you have a function \( g(x) = h(f(x)) \), and you want to find \( g'(x) \), you'll take the derivative of the outer function evaluated at the inner function \( h'(f(x)) \), and multiply it by the derivative of the inner function \( f'(x) \).
• This is expressed as \( g'(x) = h'(f(x)) \cdot f'(x) \).
• In our exercise, for example, \( u(x) = \sqrt{2x-1} = (2x-1)^{1/2} \), the outer function is \( (2x-1)^{1/2} \), and the inner function is \( 2x-1 \).
• We use the chain rule to first differentiate the outer part \( (2x-1)^{1/2} \), which is \( \frac{1}{2}(2x-1)^{-1/2} \), then multiply it by the derivative of \( 2x-1 \), which is \( 2 \).
• Resulting in \( \frac{1}{\sqrt{2x-1}} \) – a great example of the chain rule simplifying your differentiation tasks.

This rule allows us to unpack and handle the complexities of nested functions.

Power Rule

The power rule is one of the simplest and most powerful tools in differentiation. It's particularly handy when dealing with polynomials or terms raised to a power.

Here's the main idea:

• If a function \( h(x) = x^n \), the derivative \( h'(x) \) is obtained by bringing down the exponent \( n \) and multiplying it by the base raised to a power one less: \( h'(x) = nx^{n-1} \).
• In our case, when we differentiated \( v(x) = (x-1)^2 \), we applied the power rule easily by setting \( n=2 \). This gives us \( v'(x) = 2(x-1)^{1} \).
• This straightforward process makes handling polynomials particularly efficient.

Part and parcel of simplifying any derivative task is recognizing where the power rule applies. Combining it with other rules, like the quotient or chain rule, streamlines more complicated problems.