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Find the derivative of the following functions. $$y=(3 t-1)(2 t-2)^{-1}$$

Short Answer

Expert verified
Question: Determine the derivative of the function $$y=(3t-1)(2t-2)^{-1}$$ with respect to t. Answer: $$\frac{dy}{dt} = -6(3t - 1)(2t - 2)^{-2} + 3(2t - 2)^{-1}$$

Step by step solution

01

Identify u(t) and v(t)

In the given function, we have: u(t) = (3t - 1) and v(t) = (2t - 2)^{-1}.
02

Find the derivative of u(t)

To find the derivative of u(t) = (3t - 1) with respect to t, differentiate with respect to t: $$\frac{du}{dt} = \frac{d(3t-1)}{dt} = 3$$
03

Find the derivative of v(t)

To find the derivative of v(t) = (2t - 2)^{-1} with respect to t, we make use of the chain rule: The chain rule states that if we have a composite function g(f(t)), then its derivative with respect to t is: $$\frac{dg}{dt} = g'(f(t))\frac{df}{dt}$$ Here, we have g(x) = x^{-1} and f(t) = (2t - 2). So, first, we find the derivative of g(x) and f(t) with respect to their respective variables: $$g'(x) = \frac{d(x^{-1})}{dx} = -x^{-2}$$ $$\frac{df}{dt} = \frac{d(2t - 2)}{dt} = 2$$ Now, we can apply the chain rule: $$\frac{dv}{dt} = g'(f(t))\frac{df}{dt} = -f(t)^{-2} \cdot 2 = -2(2t - 2)^{-2}$$
04

Apply the product rule to find the derivative of y(t)

We can now apply the product rule formula to find the derivative of the given function: $$\frac{dy}{dt} = u\frac{dv}{dt} + v\frac{du}{dt} = (3t - 1)\left(-2(2t - 2)^{-2}\right) + (2t - 2)^{-1}(3)$$
05

Simplify the expression for dy/dt

Now, we can simplify the expression for the derivative: $$\frac{dy}{dt} = -6(3t - 1)(2t - 2)^{-2} + 3(2t - 2)^{-1}$$ This is the derivative of the given function with respect to t: $$\frac{dy}{dt} = -6(3t - 1)(2t - 2)^{-2} + 3(2t - 2)^{-1}$$

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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 tool in calculus for differentiating composite functions. A composite function is essentially a function within another function. To differentiate such functions, we can use the chain rule. Here's how it works:
  • Suppose we have two functions, say, \( g(x) \) and \( f(t) \), forming a composite function \( g(f(t)) \).
  • The derivative of \( g(f(t)) \) with respect to \( t \) is found by first taking the derivative of the outer function \( g(x) \) with respect to its own variable, resulting in \( g'(x) \).
  • Next, we multiply it by the derivative of the inner function \( f(t) \) with respect to \( t \).
In mathematical terms, this operation is represented as:\[\frac{dg}{dt} = g'(f(t)) \cdot \frac{df}{dt}\]
This rule is particularly useful when dealing with functions that look complex but are made up of simpler functions nested within each other. Understanding and practicing the chain rule can greatly enhance your calculus skills.
Product Rule
The product rule is a technique used in calculus to differentiate functions that are products of two other functions. If you have a function \( y(t) \) expressed as the product of two functions, say \( u(t) \) and \( v(t) \), the product rule applies:
  • The derivative of this product \( u(t) \times v(t) \) with respect to \( t \) is given by the sum of two products.
  • First, you take the derivative of \( u(t) \), and multiply it by \( v(t) \) without differentiating \( v(t) \).
  • Second, you add a product where you multiply \( u(t) \) without differentiating by the derivative of \( v(t) \).
This is beautifully captured by the formula:\[\frac{dy}{dt} = u(t) \frac{dv}{dt} + v(t) \frac{du}{dt}\]
The product rule is fundamental when differentiating expressions where terms are multiplied together. It ensures that every aspect of the multiplying functions contributes accurately to the derivative of the overall expression. Applying this rule allows you to handle more complex differential equations.
Differentiation
Differentiation is the process of calculating a derivative, which measures how a function changes as its input changes. A derivative is a fundamental concept in calculus, representing the slope or rate of change of a function at a given point.
  • The process involves several rules and methods, like the chain rule and product rule, to handle different kinds of functions.
  • The result of differentiation gives us a new function that describes how the original function behaves as its input varies.
In practical terms, differentiation tells us how rapidly a function is increasing or decreasing. You can think of it as determining the "velocity" of a function's output with respect to changes in its input. This concept is integral in fields ranging from physics to economics, offering insights into behaviors modeled by mathematical functions.
By mastering differentiation, you open the doors to understanding motion, growth, optimization, and much more, making it a cornerstone of mathematical analysis.

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Most popular questions from this chapter

Suppose a large company makes 25,000 gadgets per year in batches of \(x\) items at a time. After analyzing setup costs to produce each batch and taking into account storage costs, it has been determined that the total cost \(C(x)\) of producing 25,000 gadgets in batches of \(x\) items at a time is given by $$C(x)=1,250,000+\frac{125,000,000}{x}+1.5 x.$$ a. Determine the marginal cost and average cost functions. Graph and interpret these functions. b. Determine the average cost and marginal cost when \(x=5000\). c. The meaning of average cost and marginal cost here is different from earlier examples and exercises. Interpret the meaning of your answer in part (b).

Two boats leave a port at the same time, one traveling west at \(20 \mathrm{mi} / \mathrm{hr}\) and the other traveling southwest at \(15 \mathrm{mi} / \mathrm{hr} .\) At what rate is the distance between them changing 30 min after they leave the port?

The output of an economic system \(Q,\) subject to two inputs, such as labor \(L\) and capital \(K\) is often modeled by the Cobb-Douglas production function \(Q=c L^{a} K^{b} .\) When \(a+b=1,\) the case is called constant returns to scale. Suppose \(Q=1280, a=\frac{1}{3}, b=\frac{2}{3},\) and \(c=40\) a. Find the rate of change of capital with respect to labor, \(d K / d L\). b. Evaluate the derivative in part (a) with \(L=8\) and \(K=64\)

The total energy in megawatt-hr (MWh) used by a town is given by $$E(t)=400 t+\frac{2400}{\pi} \sin \frac{\pi t}{12},$$ where \(t \geq 0\) is measured in hours, with \(t=0\) corresponding to noon. a. Find the power, or rate of energy consumption, \(P(t)=E^{\prime}(t)\) in units of megawatts (MW). b. At what time of day is the rate of energy consumption a maximum? What is the power at that time of day? c. At what time of day is the rate of energy consumption a minimum? What is the power at that time of day? d. Sketch a graph of the power function reflecting the times at which energy use is a minimum or maximum.

A thin copper rod, 4 meters in length, is heated at its midpoint, and the ends are held at a constant temperature of \(0^{\circ} .\) When the temperature reaches equilibrium, the temperature profile is given by \(T(x)=40 x(4-x),\) where \(0 \leq x \leq 4\) is the position along the rod. The heat flux at a point on the rod equals \(-k T^{\prime}(x),\) where \(k>0\) is a constant. If the heat flux is positive at a point, heat moves in the positive \(x\) -direction at that point, and if the heat flux is negative, heat moves in the negative \(x\) -direction. a. With \(k=1,\) what is the heat flux at \(x=1 ?\) At \(x=3 ?\) b. For what values of \(x\) is the heat flux negative? Positive? c. Explain the statement that heat flows out of the rod at its ends.

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