Problem 11
Question
Find the value of the derivative of the function at the given point. State which differentiation rule you used to find the derivative. $$\begin{array}{ll}{\text { Function }} & {\text { Point }} \\ {f(t)=\frac{2 t^{2}-3}{3 t+1}} & {\left(3, \frac{3}{2}\right)} \end{array}$$
Step-by-Step Solution
Verified Answer
The value of the derivative of the function at the given point (3, 3/2) is 0.75. The differentiation rule used is the quotient rule.
1Step 1: Identify the Functions
Identify the numerator and denominator functions as \(f(t) = 2t^2 - 3\) and \(g(t) = 3t + 1\), respectively.
2Step 2: Apply the Quotient Rule
Compute the derivative of the function \(f(t)\) using the quotient rule: \((g(t)*f'(t) - f(t)*g'(t))/(g(t))^2\). Note that \(f'(t) = 4t\) and \(g'(t) = 3\). Substituting these into the quotient rule equation yields: \((3t+1)*4t - (2t^2-3)*3 / (3t+1)^2\). Simplifying this gives \((12t^2 + 4t - 6t^2 + 9) / (9t^2 + 6t + 1)\), or \((6t^2 + 4t + 9) / (9t^2 + 6t + 1)\).
3Step 3: Evaluate the Derivative
Evaluate the derivative at the point (3, 3/2) by substituting \(t = 3\) into the derivative function. The result is \((6*(3^2) + 4*3 + 9) / (9*(3^2) + 6*3 + 1)\) which simplifies to \((54 + 12 + 9) / (81 + 18 + 1) = 75/100 = 0.75\).
Key Concepts
Derivative of a functionDifferentiation techniquesCalculus problem solving
Derivative of a function
The derivative of a function is a fundamental concept in calculus that measures how a function changes as its input changes. In simple terms, it tells us how the value of a function is behaving at any given point, such as whether it's increasing or decreasing. The derivative itself is another function that provides information about the rate of change of the original function's values.
When we talk about finding the derivative, we're essentially looking for the slope of the tangent line at any point on a function's curve. This process is called differentiation. Knowing how to compute derivatives allows us to solve a wide array of real-world problems involving rates of change, such as speeds and growth rates.
In the given problem, finding the derivative of the function at a specific point gives us the instantaneous rate of change at that point. This can be visualized as how steep the function is at \(t=3\). Understanding the concept of derivatives opens up insights into the behavior of functions, making calculus an essential tool in both mathematics and applied sciences.
When we talk about finding the derivative, we're essentially looking for the slope of the tangent line at any point on a function's curve. This process is called differentiation. Knowing how to compute derivatives allows us to solve a wide array of real-world problems involving rates of change, such as speeds and growth rates.
In the given problem, finding the derivative of the function at a specific point gives us the instantaneous rate of change at that point. This can be visualized as how steep the function is at \(t=3\). Understanding the concept of derivatives opens up insights into the behavior of functions, making calculus an essential tool in both mathematics and applied sciences.
Differentiation techniques
Differentiation techniques are methods used to find the derivative of a function. One fundamental technique used in our problem is the Quotient Rule. This rule specifically helps in differentiating functions that are expressed as the ratio of two other functions, just like the function you've dealt with: \((f(t)=\frac{2t^{2}-3}{3t+1})\).
The Quotient Rule states that if you have a function that is \(\frac{u(t)}{v(t)}\), the derivative can be calculated using: \((v(t) \cdot u'(t) - u(t) \cdot v'(t))/(v(t))^2\).
This formula breaks down differentiation into manageable steps by isolating the numerator and denominator of the function, making it simpler to handle complex expressions. Other differentiation techniques include the product rule for products of functions and the chain rule for composite functions. Each rule in calculus serves its unique purpose, catering to different types of functions to ensure we can handle a wide variety of problems.
The Quotient Rule states that if you have a function that is \(\frac{u(t)}{v(t)}\), the derivative can be calculated using: \((v(t) \cdot u'(t) - u(t) \cdot v'(t))/(v(t))^2\).
This formula breaks down differentiation into manageable steps by isolating the numerator and denominator of the function, making it simpler to handle complex expressions. Other differentiation techniques include the product rule for products of functions and the chain rule for composite functions. Each rule in calculus serves its unique purpose, catering to different types of functions to ensure we can handle a wide variety of problems.
Calculus problem solving
Calculus problem solving involves systematically applying calculus concepts and techniques to find solutions to mathematical problems. The process generally starts with understanding the problem, identifying the functions involved, determining which differentiation rules apply, and then performing the required computations.
Approach problem-solving in calculus with these steps:
Approach problem-solving in calculus with these steps:
- Identify the function(s) you need to differentiate.
- Choose the appropriate differentiation technique, such as the Quotient Rule used here.
- Break down the problem using known rules and simplify at each step.
- Evaluate the derivative at the given point (as seen in the problem, where \(t = 3\)).
- Verify your final answer for accuracy.
Other exercises in this chapter
Problem 11
Find \(d y / d u, d u / d x,\) and \(d y / d x.\) $$ y=\sqrt{u}, u=3-x^{2} $$
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find the second derivative of the function. $$ y=\left(x^{3}-2 x\right)^{4} $$
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Use a graphing utility to graph the function and find its average rate of change on the interval. Compare this rate with the instantaneous rates of change at th
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Find the derivative of the function. $$ y=t^{2}-6 $$
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