Problem 36 - Calculus for Biology and Medicine

Question

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

Step-by-Step Solution

Answer

The derivative is \( f'(x) = \frac{4x - 1}{4x^2 - 2x} \).

Step 1: Simplify the Expression

The given function is \( f(x) = \ln \sqrt{2x^2 - x} \). We can simplify this by using the property of logarithms that \( \ln \sqrt{u} = \frac{1}{2} \ln u \). Thus, \( f(x) = \frac{1}{2} \ln (2x^2 - x) \).

Step 2: Apply the Chain Rule

To differentiate \( f(x) = \frac{1}{2} \ln (2x^2 - x) \), use the chain rule which states that if \( y = \ln u \), then \( \frac{dy}{dx} = \frac{1}{u} \cdot \frac{du}{dx} \). Here, \( u = 2x^2 - x \).

Step 3: Differentiate the Inner Function

Differentiate the inner function \( u = 2x^2 - x \). The derivative is \( \frac{du}{dx} = 4x - 1 \).

Step 4: Differentiate the Logarithmic Function

Using the results from the previous steps, the derivative of \( y = \frac{1}{2} \ln (2x^2 - x) \) is: \[ f'(x) = \frac{1}{2} \times \frac{1}{2x^2 - x} \times (4x - 1) = \frac{4x - 1}{2(2x^2 - x)}. \]

Step 5: Simplify the Derivative

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

Key Concepts

Chain Rule

The chain rule is an essential concept in calculus, particularly when differentiating composite functions. It provides a way to differentiate a function that is the composition of two or more functions. In simpler terms, when you have a function inside another function, the chain rule helps us find the derivative of the entire composition.
Let's consider our example:

• The original function is: \(f(x)=rac{1}{2} \ln(2x^2 - x)\).
• Here, \(\ln(2x^2 - x)\) is a composite function where the outer function is the natural logarithm \(\ln\) and the inner function is \(u = 2x^2 - x\).

To use the chain rule, differentiate the outer function with respect to the inner function, and then multiply that result by the derivative of the inner function with respect to \(x\). In mathematical terms, if a function \(y = \ln(u)\), the derivative \(\frac{dy}{dx}\) is \(\frac{1}{u} \cdot \frac{du}{dx}\). This step helps us break down complex derivatives into simpler, manageable parts.

Logarithmic Differentiation

Logarithmic differentiation is a technique used especially when differentiating functions that involve products, quotients, or powers, making them into simpler forms using logarithms. In our case, while \(f(x)=\ln\sqrt{2x^2 - x}\), we simplify it initially using the property of logarithms: \(\ln(\sqrt{u}) = \frac{1}{2} \ln(u)\). This sets the function as \(f(x) = \frac{1}{2} \ln(2x^2 - x)\).
This manipulation allows us to straightforwardly apply the chain rule for an easier differentiation process, as it turns the composite inside square roots into logarithmic form which is simpler to handle mathematically. Logarithmic differentiation is powerful because it can handle forms that are otherwise challenging using standard differentiation rules.

Derivative Simplification

Once the differentiation is complete, it's crucial to simplify the derivative, both for an accurate and easily interpretable result. In our example:

• We find the derivative formula: \(f'(x) = \frac{4x - 1}{2(2x^2 - x)}\).
• This becomes \(f'(x) = \frac{4x - 1}{4x^2 - 2x}\) upon simplification.

Simplification involves breaking down the results into the simplest form possible. To simplify:

• Factor out common elements in the numerator and denominator where possible.
• You may need to cancel out terms, combine like terms, or simplify fractions.

Effective simplification not only makes the final answer neat but also reveals deeper insights into the behavior of the function and its derivatives. Moreover, it often prevents computational errors in further calculations as it reduces the complexity of the expressions involved.