Problem 44 - Calculus and its applications

Question

Differentiate each function. \(G(t)=\frac{1}{t+2}\)

Step-by-Step Solution

Answer

The derivative of \(G(t)=\frac{1}{t+2}\) is \(G'(t)=-\frac{1}{(t+2)^2}\).

Step 1: Recall the Differentiation Rule for Division

To differentiate the function \(G(t)=\frac{1}{t+2}\), recognize it is in the form of a quotient where the numerator is a constant (1) and the denominator is a linear function \(t+2\). The derivative of a constant divided by a function can be expressed using the formula for differentiation of \(f(x)=\frac{1}{g(x)}\), which is \(f'(x) = -\frac{g'(x)}{(g(x))^2}\).

Step 2: Differentiate the Denominator

Identify the function \(g(t)=t+2\). Compute its derivative: \(g'(t)=1\). This means that the rate at which \(t+2\) changes with respect to \(t\) is constant at 1.

Step 3: Apply the Differentiation Formula

Substitute \(g(t)=t+2\) and \(g'(t)=1\) into the differentiation formula: \(G'(t) = -\frac{g'(t)}{(g(t))^2}\). This gives \(G'(t) = -\frac{1}{(t+2)^2}\).

Step 4: Simplify the Expression

Since the formula directly gives us the derivative, there is no more simplification needed. The final derivative is \(G'(t) = -\frac{1}{(t+2)^2}\).

Key Concepts

Quotient Rule

The quotient rule is a fundamental method in calculus used to differentiate functions that are ratios of two separate functions. It helps us to calculate the derivative of expressions like \( G(t) = \frac{1}{t+2} \), where one function is divided by another.

For functions in the form \( f(x) = \frac{u(x)}{v(x)} \), the quotient rule states that the derivative \( f'(x) \) is given by:
\[ f'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{(v(x))^2} \]

In essence, the rule combines the derivatives of the numerator and the denominator using a systematic formula to account for their individual rates of change. It is crucial to distinguish the roles of the numerator and denominator and correctly apply the formula to compute the derivative without errors.

Linear Functions

Linear functions are the simplest form of polynomial functions. They are characterized by their straight-line graph and have the general form \( f(x) = mx + c \), where \( m \) and \( c \) are constants.

In the context of differentiation, linear functions are particularly interesting because their derivatives are constant. For example, given a linear function \( g(t) = t + 2 \), its derivative \( g'(t) = 1 \).

Key characteristics of linear functions include:

• A constant slope \( m \), which represents the rate of change of the function.
• A y-intercept \( c \), which is the value of the function when \( x = 0 \).

These properties make linear functions straightforward to differentiate, simplifying the process and serving as ideal building blocks for understanding more complex calculus problems.

Derivative Calculation

Calculating derivatives is a core practice in calculus, enabling us to understand how functions change at any point. In the case of \( G(t)=\frac{1}{t+2} \), the derivative calculates the rate of change of the function with respect to \( t \).

Here, we apply the formula for the derivative of a reciprocal function, \( f(x) = \frac{1}{g(x)} \), which is:
\[ f'(x) = -\frac{g'(x)}{(g(x))^2} \]
By following these steps:

• Identify \( g(t) = t+2 \) and calculate \( g'(t) = 1 \).
• Substitute into the formula to get \( G'(t) = -\frac{1}{(t+2)^2} \).

This calculation shows that the derivative of \( G(t) \) at any point \( t \) is the negative reciprocal of the square of the original function's denominator, indicating how sharply it decreases.