Question

Using the Quotient Rule In Exercises \(7-12,\) use the Quotient Rule to find the derivative of the function. $$ g(t)=\frac{3 t^{2}-1}{2 t+5} $$

Short answer

The derivative of the function, \(g(t) = \frac{{3t^{2}-1}}{{2t+5}}\), is \(g'(t) = \frac{{9t}}{{(2t+5)^2}}\).

Step-by-step solution

Identify The Functions

Identify the numerator function \(f(t) = 3t^{2}-1\), and the denominator function \(h(t) = 2t+5\), from the given function: \(g(t) = \frac{{f(t)}}{{h(t)}}\). Taking derivatives gives \(f'(t) = 6t\) and \(h'(t) = 2\).

Apply The Quotient Rule

Apply the Quotient Rule, which states: \(g'(t) = \frac{{f'(t)h(t) - f(t)h'(t)}}{{[h(t)]^2}}\).

Substitute The Functions And Their Derivatives Into The Equation

Substitute \(f(t), f'(t), h(t),\) and \(h'(t)\) into the equation to compute \(g'(t) = \frac{{6t(2t+5) - (3t^{2}-1) * 2}}{{(2t+5)^2}}\).

Simplify The Equation

Simplify the equation to get the final derivative function and we get \(g'(t) = \frac{{9t}}{{(2t+5)^2}}\).

Key concepts

Derivative of a Function

Understanding the derivative of a function is a vital concept in calculus. It signifies how a function's output changes as its input changes. Loosely speaking, a derivative represents the slope of the function at a given point on its graph; it shows the rate at which the function's value is changing at that point.

For instance, if distance is a function of time, the derivative would be the speed, indicating how quickly the distance is traversing as time advances. In the exercise, we are tasked with finding the derivative of the function using the Quotient Rule, which essentially calculates the rate of change of one quantity relative to another.

Let's consider the exercise's example, where we have the function \( g(t) = \frac{3 t^{2}-1}{2 t+5} \). Here, the numerator (\( f(t) = 3t^{2}-1 \) represents one quantity, and the denominator (\( h(t) = 2t+5 \) represents another. After determining their derivatives, which are \( f'(t) = 6t \) and \( h'(t) = 2 \) respectively, the Quotient Rule is applied to find the derivative of the function given by \( g(t) \).

Calculus

Calculus is the mathematical study of continuous change. It provides us with the tools for modeling and solving problems where variables change dynamically. There are two fundamental branches of calculus: differential calculus (concerning the derivative) and integral calculus (concerning the antiderivative or integral).

In the context of the exercise, calculus is used to determine how functions like \( g(t) \) change over the variable \( t \). Through calculus, we can find the instantaneous rate of change, area under curves, and the accumulated quantity based on the variable function's rate. The process we use in the exercise, differentiation, is a major workhorse of calculus, enabling us to handle complex relationships between rapidly changing quantities.

Differentiation

Differentiation is the process of finding the derivative of a function. It's one of the central operations in calculus. When differentiating functions, we have several rules to assist us, such as the Power Rule, Product Rule, Chain Rule, and the Quotient Rule as seen in the exercise.

In using the Quotient Rule, we differentiate a function of the form \( \frac{f(t)}{h(t)} \) where \( f(t) \) and \( h(t) \) are both differentiable functions of \( t \). The rule states that the derivative of the function is \( g'(t) = \frac{f'(t)h(t) - f(t)h'(t)}{[h(t)]^2} \). This rule is essentially an algorithm for division in the language of calculus.

The application of the Quotient Rule in the exercise clearly demonstrated the step-by-step approach to differentiation when dealing with a ratio of functions. By meticulously following the steps and simplifying, we obtained the derivative \( g'(t) = \frac{9t}{(2t+5)^2} \), expressing how the function \( g(t) \) changes at any point \( t \).