Problem 33 - Calculus: Early Transcendentals

Question

Find \( f'(a) \). \( f(t) = \dfrac{2t + 1}{t + 3} \)

Step-by-Step Solution

Answer

The derivative is \( f'(a) = \frac{5}{(a + 3)^2} \).

Step 1: Recall the Quotient Rule

To differentiate a function that is the quotient of two functions, \( f(t) = \frac{u(t)}{v(t)} \), we use the Quotient Rule. The Quotient Rule states that \( (\frac{u}{v})' = \frac{u'v - uv'}{v^2} \).

Step 2: Identify Functions and Derivatives

Here, let \( u(t) = 2t + 1 \) and \( v(t) = t + 3 \). The derivatives are \( u'(t) = 2 \) and \( v'(t) = 1 \).

Step 3: Apply the Quotient Rule

Using the quotient rule, we compute: \[ \left( \frac{2t + 1}{t + 3} \right)' = \frac{(2)(t + 3) - (2t + 1)(1)}{(t + 3)^2} \] This gives us: \[ \frac{2t + 6 - 2t - 1}{(t + 3)^2} = \frac{5}{(t + 3)^2} \]

Step 4: Evaluate the Derivative at \( t = a \)

Now we substitute \( t = a \) into the derivative: \[ f'(a) = \frac{5}{(a + 3)^2} \].

Key Concepts

Quotient Rule

The Quotient Rule is a method used in calculus for differentiating functions that are expressed as ratios of two simpler functions. When you see a function in the form \( f(t) = \frac{u(t)}{v(t)} \), the Quotient Rule helps to find its derivative without much hassle. Here's a quick way to remember it:

• "Low d-high minus high d-low, square the bottom, and away we go!"

This means you multiply the derivative of the numerator "high" by the denominator "low", and subtract the product of the numerator and the derivative of the denominator. Then you run all over the square of the denominator. Mathematically, it's expressed as:\[\left(\frac{u}{v}\right)' = \frac{u'v - uv'}{v^2}\]Using this rule simplifies the differentiation process of ratio functions, making it straightforward to handle more complex equations with confidence.

Differentiation

Differentiation is one of the core concepts of calculus that deals with finding the derivative of a function. The derivative essentially represents the rate at which a function is changing at any given point. Think of it as determining how steep a hill is, at any particular slope on its curve.

In mathematical terms, the derivative of a function \( f(t) \) at any point \( t \) describes the instantaneous rate of change of the function with respect to \( t \). If you picture a curve on a graph, the derivative tells you the slope of the tangent line at any point on that curve.

• For linear functions, the derivative is a single value as the slope is constant.
• For more complex functions, the derivative becomes an expression representing that change.

Differentiation rules such as the Quotient Rule, Product Rule, and Chain Rule help in dealing with diverse mathematical equations, ensuring you get the right derivative every time.

Functions

Functions are mathematical entities that link an input to an output in a consistent way. In simpler terms, you can think of a function as a machine that churns out a unique result whenever you feed it a particular input. Each function has its own rule or formula that defines the relationship between the input and output.

• An example of a function is \( f(t) = \frac{2t + 1}{t + 3} \), which takes an input \( t \) and transforms it according to the equation's rule to produce an output.

In calculus, understanding functions and their behavior is crucial because they form the basis for differentiation and integration. By grasping the nature of functions, you can better understand how to apply differentiation rules to find the slope or rate of change, as seen with the Quotient Rule used in differentiating the provided example. Functions can vary widely in complexity, but the foundational idea remains focused around generating an output based on given inputs, whether those are numbers, vectors, or other mathematical objects.