Problem 13 - Calculus an Applied Approach

Question

Find the value of the derivative of the function at the given point. State which differentiation rule you used to find the derivative. $$\begin{array}{ll}{\text { Function }} & {\text { Point }} \\\\{g(x)=\frac{2 x+1}{x-5}} & {(6,13)} \end{array}$$

Step-by-Step Solution

Answer

The derivative of the function \(g(x) = \frac{2x + 1}{x - 5}\) at the point (6,13) is -3. The quotient rule of differentiation was used to find this derivative.

Step 1: Identify u, v, u' and v'

Identify functions \(u\) and \(v\) and calculate their respective derivatives \(u'\) and \(v'\). Here, \(u = 2x + 1\) so \(u' = 2\), and \(v = x - 5\) so \(v' = 1\).

Step 2: Apply the quotient rule

Apply the quotient rule to calculate the derivative of \(g(x)\). This gives \(g'(x) = \frac{(x - 5) * 2 - (2x + 1) * 1}{(x - 5)^2}\). Simplify this to get \(g'(x) = \frac{-3}{(x-5)^2}\).

Step 3: Substitute the given x-value

Now, substitute \(x = 6\) into the derivative to find its value at this point. This results in \(g'(6) = -3\).

Key Concepts

Quotient Rule

When you are confronted with finding the derivative of a function that is a ratio of two functions, you use the quotient rule. This is particularly useful in calculus differentiation. The quotient rule helps us to differentiate larger and more complex functions, composed of a numerator function and a denominator function. Here is the structure of the quotient rule formula:

• If \( u(x) \) and \( v(x) \) are functions, then the derivative of their quotient \( \frac{u}{v} \) is given by:
\[ \left( \frac{u}{v} \right)' = \frac{u'v - uv'}{v^2} \]
• "u'" and "v'" are the derivatives of "u" and "v" respectively. The quotient rule formula shows us that it involves multiplying, subtracting, and dividing components of the given functions.

This rule is particularly useful when directly applying other differentiation methods might seem challenging or less intuitive.

Derivative Calculation

Calculating derivatives involves using certain rules to find the derivative, which is the rate of change of a function. In our exercise, we calculated the derivative using the quotient rule. It is important to perform this correctly in steps:

• First, identify your numerator function, \( u(x) \), and your denominator function, \( v(x) \).
• Calculate the derivatives of both functions separately to get \( u' \) and \( v' \).
• Substitute these into the quotient rule formula \( \frac{u'v - uv'}{v^2} \) to find the derivative of the quotient.
• Make sure all terms are simplified for a clean and correct derivative expression.

Understanding each step clearly will greatly improve your ability to work through derivative problems with confidence.

Function Derivatives

A function derivative explains the rate at which a function changes. It's like finding the function's speed at any given point. For the function \( g(x) = \frac{2x+1}{x-5} \), we know this rate of change can tell us how steeply the function is rising or falling.

• The initial step involves identifying the function's components and their individual derivatives.
• These derivatives, \( u' \) and \( v' \), are crucial in applying the quota The function \( u(x) = 2x + 1 \) has a derivative \( u' = 2 \), which tells us how quickly it changes. The function \( v(x) = x - 5 \) has a derivative \( v' = 1 \).
• The derivative of the entire function, \( g'(x) \), brings together these rates of change to give us a comprehensive view of the function's behavior.

Mastering how to find derivatives of different function types strengthens your problem-solving skills across calculus.

Finding Derivatives at a Point

Once you have found a function's derivative, you can determine the rate of change at any specific point by substituting the value for that point into your derivative.

• For example, in our original problem, after finding \( g'(x) \), we substituted \( x = 6 \) to calculate \( g'(6) \).
• This gives a numerical value indicating how the function is changing at the exact point \( x = 6 \).
• In our case, substituting gives \( g'(6) = -3 \), which means at \( x = 6 \), the function decreases at a rate of 3 units per unit increase in x.

Finding the derivative at a point is a fundamental technique, offering insights into the function's behavior at specific moments.