Question

Find the value of the derivative of the function at the given point. State which differentiation rule you used to find the derivative. $$ f(t)=\frac{2 t^{2}-3}{3 t+1} \quad\left(3, \frac{3}{2}\right) $$

Short answer

The value of the derivative at the point (3, 3/2) is 0.75. The rule used to find the derivative was the Quotient Rule of differentiation.

Step-by-step solution

Identify the function forms

The given function \(f(t)=\frac{2 t^{2}-3}{3 t+1}\) is a quotient of two functions (a numerator and a denominator). So, the quotient rule of differentiation will be used.

Apply the Quotient Rule

The quotient rule of differentiation states that the derivative of a quotient \(\frac{u}{v}\) is \(\frac{v(u') - u(v')}{v^{2}}\). Here \(u = 2t^2 - 3\) and \(v = 3t + 1\), so their derivatives are \(u' = 4t\) and \(v' = 3\). The derivative of \(f(t)\) will thus be \(f'(t) = \frac{(3t + 1)(4t) - (2t^2 - 3)(3)}{(3t + 1)^2}\).

Simplify the Derivative

By simplifying the numerator and the denominator separately, we obtain \(f'(t) = \frac{12t^2 + 4t - 6t^2 + 9}{(3t + 1)^2} = \frac{6t^2 + 4t + 9}{(3t + 1)^2}\).

Evaluate the derivative at the given point

Now we substitute the coordinates of the given point \(t = 3\) into the derivative to find its value at that point: \(f'(3) = \frac{6*3^2 + 4*3 + 9}{(3*3 + 1)^2} = \frac{6*9 + 12 + 9}{10^2} = \frac{54 + 12 + 9}{100} = \frac{75}{100} = 0.75\).

Key concepts

Derivative Calculation

Understanding how to calculate derivatives is fundamental in calculus. It's the process of finding the rate at which a function is changing at any point. For the function
\( f(t)=\frac{2t^2-3}{3t+1} \), we're dealing with a quotient of two functions. The quotient rule is a precise method for tackling such a calculation.
To grasp how it works, think of a function divided by another as an elegant dance where each participant must be acknowledged. The quotient rule states that if you have a function \( u \) divided by a function \( v \), the derivative \( f'(t) \) is \( \frac{vu' - uv'}{v^2} \), where \( u' \) and \( v' \) are the derivatives of \( u \) and \( v \), respectively. This is how we choreograph the complex movement of our function's rate of change.

Applying Differentiation Rules

Differentiation rules are like tools in a toolbox, each with a specific use. For different function forms—whether they're polynomials, trigonometric, or like ours, a quotient—there's an appropriate rule.
The quotient rule is for when one function is divided by another. Applying it correctly requires us to differentiate the numerator and denominator separately, which is what we do with our functions \( u = 2t^2 - 3 \) and \( v = 3t + 1 \). Afterward, we combine these derivatives following the quotient rule formula. Missteps in application could lead to incorrect results, hence why correctly identifying the rule needed and applying it stepwise is crucial for success in calculus.

Simplifying Derivatives

Simplifying a derivative often reveals its true nature, much like clearing clouds to reveal a sunny sky. It's not just about aesthetics; a simplified derivative is typically easier to work with and understand.
In our problem, after applying the quotient rule, simplification involves expanding the products in the numerator and then combining like terms. This trimming process, although it might seem laborious, can dramatically reduce the complexity of the expression, converting \( f'(t) = \frac{12t^2 + 4t - 6t^2 + 9}{(3t + 1)^2} \) into the more wieldy \( \frac{6t^2 + 4t + 9}{(3t + 1)^2} \). Simplicity, in this case, paves the way for smoother further calculations, like evaluating the derivative at a specific point.

Evaluating Derivatives at a Point

The act of evaluating a derivative at a point can be likened to taking a snapshot of the rate of change of a function at a particular moment. It answers the question, 'How fast is the function changing right here and now?'.
For the given function \( f(t) \), after simplification, we substitute the value \( t = 3 \) into our derivative to find \( f'(t) \) at that specific point. It is a matter of plugging the number into the simplified derivative and doing the arithmetic. This final figure, like the 0.75 we get for our function, tells us precisely the slope of the tangent line to the curve of \( f(t) \) at the point \( (3, \frac{3}{2}) \). Understanding this concept is critical because it provides insight into the function's behavior and can help predict future trends or find optimum values.