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Find \(d y / d t\) $$y=\left(\frac{t^{2}}{t^{3}-4 t}\right)^{3}$$

Short Answer

Expert verified
Apply the chain rule and quotient rule to find \(\frac{dy}{dt}\). The resulting derivative is complex and can be simplified.

Step by step solution

01

Identify the Differentiation Rule

In this problem, we are dealing with a composite function, which requires the use of both the chain rule and the quotient rule for differentiation. Recognizing that we have a function raised to a power of 3 and an internal quotient, we need to apply the chain rule followed by the quotient rule.
02

Apply the Chain Rule

The chain rule is expressed as: if you have a function \(y = u^n\), the derivative \(d y / d t\) is \(n \cdot u^{n-1} \cdot \frac{du}{dt}\). Here, \(u = \left(\frac{t^{2}}{t^{3}-4t}\right)\) and \(n = 3\). Thus, \(\frac{dy}{dt} = 3 \cdot \left(\frac{t^2}{t^3 - 4t}\right)^2 \cdot \frac{d}{dt}\left(\frac{t^2}{t^3 - 4t}\right)\).
03

Apply the Quotient Rule

To find \(\frac{du}{dt}\), apply the quotient rule given by \(\frac{d}{dt}\left(\frac{f(t)}{g(t)}\right) = \frac{f'(t)g(t) - f(t)g'(t)}{[g(t)]^2}\). Here, \(f(t) = t^2\) and \(g(t) = t^3 - 4t\). Compute the derivatives: \(f'(t) = 2t\) and \(g'(t) = 3t^2 - 4\).
04

Substitute into the Quotient Rule

Substitute \(f(t), f'(t), g(t),\) and \(g'(t)\) into the quotient rule:\[ \frac{du}{dt} = \frac{2t(t^3 - 4t) - t^2(3t^2 - 4)}{(t^3 - 4t)^2}\]Simplify the numerator:\[= \frac{2t^4 - 8t^2 - 3t^4 + 4t^2}{(t^3 - 4t)^2} = \frac{-t^4 - 4t^2}{(t^3 - 4t)^2}\]
05

Complete the Differentiation

Now substitute \(\frac{du}{dt}\) back into the chain rule expression:\[ \frac{dy}{dt} = 3 \cdot \left(\frac{t^2}{t^3 - 4t}\right)^2 \cdot \frac{-t^4 - 4t^2}{(t^3 - 4t)^2}\]This expression gives the derivative, although simplification can be done for further reduction.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Chain Rule
The chain rule is an essential differentiation technique in calculus. It helps us find the derivative of a composite function, which is a function within another function. In simpler terms, picture it like layers of an onion where you need to peel back each layer to get to the center.

The chain rule formula can be expressed as follows: if you have a function written as \( y = u^n \), then the derivative \( \frac{dy}{dt} \) is found using \( n \cdot u^{n-1} \cdot \frac{du}{dt} \). In the original exercise, \( y=\left(\frac{t^{2}}{t^{3}-4t}\right)^{3} \), we identify "\( u = \frac{t^{2}}{t^{3}-4t} \)" and an outer function raised to the power 3. Therefore, the chain rule kicks in to help differentiate it systematically.

When applying the chain rule, you need to identify these functions and their respective inner functions. This understanding will guide you through the differentiation process step by step.
Quotient Rule
The quotient rule is implemented when you have one function divided by another function. It allows you to differentiate ratios of two functions effectively, like dividing apples by oranges. Imagine sorting fruits, where you must maintain the balance between apples and oranges to reach a solution.

The rule is given by: \( \frac{d}{dt}\left(\frac{f(t)}{g(t)}\right) = \frac{f'(t)g(t) - f(t)g'(t)}{[g(t)]^2} \). Breaking it down:
  • Identify \( f(t) \) and \( g(t) \).
  • Compute their derivatives, \( f'(t) \) and \( g'(t) \).
  • Substitute these into the formula to find \( \frac{du}{dt} \).
In the given problem, \( f(t) = t^2 \) and \( g(t) = t^3 - 4t \). After calculating their derivatives, \( f'(t) = 2t \) and \( g'(t) = 3t^2 - 4 \), you will integrate them into the quotient rule. This will yield the derivative of the inside function, keeping it all aligned for the bigger differentiation task with the chain rule.
Composite Functions
Composite functions are combinations where one function is placed inside another. This creates a multilayered structure that you need to unravel to find the derivative. Think of it like a digital code where one layer encrypts another layer. The goal is to decrypt the whole sequence.

In mathematics, dealing with composite functions involves nesting expressions. For the function in our exercise, written as \( y=\left(\frac{t^{2}}{t^{3}-4t}\right)^{3} \), the outermost function is raising to the third power, while the inside function is a quotient. Recognizing these layers helps choose the appropriate rules—such as the chain rule and quotient rule—to differentiate the composite function efficiently.

Understanding composite functions equips you with the skills needed to break down complex expressions into manageable parts. This understanding allows you to tackle more advanced calculus problems, turning what initially seems daunting into a structured approach.

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Most popular questions from this chapter

Give the position function \(s=f(t)\) of an object moving along the \(s\) -axis as a function of time \(t\). Graph \(f\) together with the velocity function \(v(t)=d s / d t=f^{\prime}(t)\) and the acceleration function \(a(t)=d^{2} s / d t^{2}=f^{\prime \prime}(t) .\) Comment on the object's behavior in relation to the signs and values of \(v\) and \(a .\) Include in your commentary such topics as the following: a. When is the object momentarily at rest? b. When does it move to the left (down) or to the right (up)? c. When does it change direction? d. When does it speed up and slow down? e. When is it moving fastest (highest speed)? Slowest? f. When is it farthest from the axis origin? $$s=t^{3}-6 t^{2}+7 t, \quad 0 \leq t \leq 4$$

Find the value of \((f \circ g)^{\prime}\) at the given value of \(x\). $$f(u)=\left(\frac{u-1}{u+1}\right)^{2}, \quad u=g(x)=\frac{1}{x^{2}}-1, \quad x=-1$$

Give the position function \(s=f(t)\) of an object moving along the \(s\) -axis as a function of time \(t\). Graph \(f\) together with the velocity function \(v(t)=d s / d t=f^{\prime}(t)\) and the acceleration function \(a(t)=d^{2} s / d t^{2}=f^{\prime \prime}(t) .\) Comment on the object's behavior in relation to the signs and values of \(v\) and \(a .\) Include in your commentary such topics as the following: a. When is the object momentarily at rest? b. When does it move to the left (down) or to the right (up)? c. When does it change direction? d. When does it speed up and slow down? e. When is it moving fastest (highest speed)? Slowest? f. When is it farthest from the axis origin? $$s=t^{2}-3 t+2, \quad 0 \leq t \leq 5$$

Find \(d y / d t\) $$y=(1+\cot (t / 2))^{-2}$$

In the late 1860 s, Adolf Fick, a professor of physiology in the Faculty of Medicine in Würzberg, Germany, developed one of the methods we use today for measuring how much blood your heart pumps in a minute. Your cardiac output as you read this sentence is probably about \(7 \mathrm{L} / \mathrm{min} .\) At rest it is likely to be a bit under \(6 \mathrm{L} / \mathrm{min}\). If you are a trained marathon runner running a marathon, your cardiac output can be as high as \(30 \mathrm{L} / \mathrm{min}\). Your cardiac output can be calculated with the formula $$y=\frac{Q}{D},$$ where \(Q\) is the number of milliliters of \(\mathrm{CO}_{2}\) you exhale in a minute and \(D\) is the difference between the \(\mathrm{CO}_{2}\) concentration \((\mathrm{mL} / \mathrm{L})\) in the blood pumped to the lungs and the \(\mathrm{CO}_{2}\) concentration in the blood returning from the lungs. With \(Q=233 \mathrm{mL} / \mathrm{min}\) and \(D=97-56=41 \mathrm{mL} / \mathrm{L}\), \(y=\frac{233 \mathrm{mL} / \mathrm{min}}{41 \mathrm{mL} / \mathrm{L}} \approx 5.68 \mathrm{L} / \mathrm{min}\), fairly close to the \(6 \mathrm{L} / \mathrm{min}\) that most people have at basal (resting) conditions. (Data courtesy of J. Kenneth Herd, M.D., Quillan College of Medicine, East Tennessee State University.) Suppose that when \(Q=233\) and \(D=41,\) we also know that \(D\) is decreasing at the rate of 2 units a minute but that \(Q\) remains unchanged. What is happening to the cardiac output?

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