Question

Find \(f^{\prime}(x)\) for each function. $$ f(x)=\ln \sqrt{5 x-7} $$

Short answer

The derivative is \(f'(x) = \frac{5}{2(5x-7)}\).

Step-by-step solution

Rewrite the Function using Logarithmic Identity

First, note that \(f(x) = \ln \sqrt{5x-7}\) can be rewritten using the logarithmic identity \(\ln a^b = b \ln a\). Hence, \(f(x) = \frac{1}{2}\ln(5x-7)\) because \(\sqrt{5x-7} = (5x-7)^{1/2}\).

Apply the Chain Rule

To differentiate \(f(x) = \frac{1}{2} \ln(5x-7)\), use the chain rule. Let \(u = 5x - 7\), then \(f(x) = \frac{1}{2} \ln(u)\). The derivative \(\frac{d}{du}(\ln(u)) = \frac{1}{u}\). Thus, \(\frac{d}{dx}(\ln(u)) = \frac{1}{u} \cdot \frac{du}{dx}\).

Differentiate the Inner Function

Find \(\frac{du}{dx}\), where \(u = 5x - 7\). The derivative is simply \(\frac{d}{dx}(5x-7) = 5\).

Combine Derivatives using the Chain Rule

Now combine the results using the chain rule: \(f'(x) = \frac{1}{2} \cdot \frac{1}{u} \cdot \frac{du}{dx} = \frac{1}{2} \cdot \frac{1}{5x-7} \cdot 5\). This simplifies to \(f'(x) = \frac{5}{2(5x-7)}\).

Key concepts

Chain Rule

The Chain Rule is an essential tool in calculus used when you need to differentiate composite functions. Imagine differentiating a function like \( f(x) = \ln \sqrt{5x-7} \). Here, you have an outer function, the natural logarithm, and an inner function, \( \sqrt{5x-7} \). The Chain Rule helps you differentiate such functions efficiently by breaking them down into simpler parts.

When applying the Chain Rule in our example, you first identify the outer and inner functions. Let the inner function be \( u = 5x - 7 \), which leads to the updated function \( f(x) = \frac{1}{2} \ln(u) \). Here’s how the Chain Rule operates:

• Differentiates the outer function with respect to the inner function: \( \frac{d}{du}(\ln(u)) = \frac{1}{u} \)
• Then multiplies by the derivative of the inner function with respect to \( x \): \( \frac{du}{dx} \)

Applying this together, your task becomes much more manageable, enabling you to get \( f'(x) = \frac{5}{2(5x-7)} \) in this case.

Logarithmic Differentiation

Logarithmic differentiation simplifies the process of differentiating complex functions, especially those involving products, quotients, or powers such as in our example. This technique takes advantage of properties of logarithms.

In the example given: \( f(x) = \ln \sqrt{5x-7} \), we use the property that \( \ln a^b = b \ln a \). This allows \( \sqrt{5x-7} \) to be rewritten as \( (5x-7)^{1/2} \), and consequently expressed as \( \frac{1}{2}\ln(5x-7) \).

By converting the original function's expression this way, the derivative becomes easier to handle, especially when coupled with the chain rule. This strategic rewriting of logarithmic expressions into simpler formats is a handy algebraic trick that can save time and simplify various differentiations.

Function Notation

Understanding function notation is crucial as it provides a clear and specific way of representing mathematical functions. In our example, function notation like \( f(x) = \ln \sqrt{5x-7} \) allows mathematical operations to be communicated precisely.

Here, \( f(x) \) signifies a function named \( f \) of the variable \( x \). The notation provides a compact way to express complex mathematical combinations, which is particularly valuable in calculus where functions often get very involved.

• The parent function \( \ln x \) is elevated by the presence of \( \sqrt{5x-7} \) as its input.
• Function notation also aids in the systematic method of differentiation, allowing each part of a compound expression to be clearly identified and worked with.

This notation remains consistent across mathematical operations, making complex derivatives follow a structured process.

Differentiation Steps

Differentiation involves a set of sequential steps to find the rate of change of a function. For the function \( f(x) = \ln \sqrt{5x-7} \), the process is broken down in a logical, ordered manner to ensure accuracy and clarity.

Here's a step-by-step breakdown:

• Step 1: Rewrite the function using logarithmic identities to simplify it. This prior simplification makes subsequent differentiation more straightforward.
• Step 2: Apply the Chain Rule. Identify the inner function \( u = 5x - 7 \) to differentiate the outer function \( \ln(u) \) initially.
• Step 3: Find the derivative of the inner function: \( \frac{du}{dx} = 5 \).
• Step 4: Combine these results using the Chain Rule to obtain the final derivative: \( f'(x) = \frac{5}{2(5x-7)} \).

Following such clear steps is important to systematically tackle more complex calculus problems with ease.