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Find the indicated derivatives. $$\frac{d v}{d t} \quad \text { if } \quad v=t-\frac{1}{t}$$

Short Answer

Expert verified
The derivative is \( \frac{d v}{d t} = 1 + \frac{1}{t^2} \).

Step by step solution

01

Identify the Function and Derivative Variable

The function given is \( v(t) = t - \frac{1}{t} \). We need to find the derivative of this function with respect to \( t \), which is denoted by \( \frac{d v}{d t} \).
02

Differentiate Each Term Separately

Differentiate the term \( t \) with respect to \( t \). The derivative is 1 since the derivative of \( t \) with respect to itself is always 1. For the second term, \( -\frac{1}{t} \), rewrite it as \( -t^{-1} \) to make differentiation easier.
03

Apply Power Rule to the Second Term

The power rule for derivatives \( \frac{d}{dt}[t^n] = n \cdot t^{n-1} \) is used. So, the derivative of \( -t^{-1} \) becomes \[ \frac{d}{dt}[-t^{-1}] = -(-1)t^{-2} = \frac{1}{t^2}. \]
04

Combine the Derivatives

Add together the derivatives of each term: \( \frac{d}{dt}[t] = 1 \) and the derivative of \( -\frac{1}{t} \) which is \( \frac{1}{t^2} \). Thus, the derivative \( \frac{d v}{d t} = 1 + \frac{1}{t^2}. \)

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Key Concepts

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

Power Rule
In calculus, one of the most fundamental techniques for finding derivatives, which measure how fast a function is changing, is known as the power rule. This rule is particularly useful for polynomials and any terms where a variable is raised to a power. It states that if you have a function in the form of \[ x^n \], where \( n \) is any real number, the derivative is\[ \frac{d}{dx} [x^n] = n \cdot x^{n-1}. \]The power rule simplifies the process of differentiation, turning potentially complicated expressions into easier forms.

Why Use the Power Rule?
  • It's fast and straightforward for terms like \( t^3 \) or \( x^7 \).
  • The power rule only requires simple multiplication and subtraction, keeping errors to a minimum.
  • It applies to any real power, even negative or fractional powers, expanding its utility across various functions.
To see the power rule in action, consider the term \( -\frac{1}{t} \), which can be rewritten as \( -t^{-1} \). Using the power rule, \[-(-1) \times t^{-1-1} = \frac{1}{t^2},\]we can find its derivative in a straightforward manner.
Derivative of a Function
The derivative of a function represents the rate at which the function's value changes as its input changes. In simpler terms, it gives you the slope of the function at any given point. If you picture the graph of a function, the derivative gives you the slope of the tangent line at any instant.

How to Find Derivatives
  • Identify the function and determine the variable with respect to which you will differentiate.
  • Apply appropriate rules of differentiation, like the power rule, to each term of the function.
  • Simplify the expression to arrive at the final derivative.
In the exercise provided, the function is \( v(t) = t - \frac{1}{t} \), and we find the derivative \( \frac{d v}{d t} \) by differentiating each term. Differentiating \( t \) relative to \( t \) gives us a derivative of \( 1 \), as any variable differentiated by itself results in 1. The key is to handle each term separately and carefully combine the results.
Calculus Problem Solving
Calculus problem-solving often involves using derivatives to analyze and understand changes in functions. Solving these problems typically requires a clear understanding of concepts and applying the right rules of calculus.

Steps to Approach Calculus Problems
  • Start by identifying the given function and the variable of differentiation.
  • Understand which differentiation rules to apply, such as the power rule or product rule.
  • Work through each term individually, applying the appropriate rule for each.
  • Combine your results to get the complete derivative of the function.
In our example, after defining the function \( v(t) = t - \frac{1}{t} \), we used the power rule to differentiate the second term \( -t^{-1} \), leading to the final derivative expression \( \frac{d v}{d t} = 1 + \frac{1}{t^2}. \)

Remember, practice and careful application of these rules is key to mastering calculus problem-solving.

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Most popular questions from this chapter

Use a CAS to estimate the magnitude of the error in using the linearization in place of the function over a specified interval \(I\). Perform the following steps: a. Plot the function \(f\) over \(I\) b. Find the linearization \(L\) of the function at the point \(a\) c. Plot \(f\) and \(L\) together on a single graph. d. Plot the absolute error \(|f(x)-L(x)|\) over \(I\) and find its maximum value. e. From your graph in part (d), estimate as large a \(\delta>0\) as you can, satisfying \(|x-a|<\delta \quad \Rightarrow \quad|f(x)-L(x)|<\epsilon\) for \(\epsilon=0.5,0.1,\) and \(0.01 .\) Then check graphically to see if your \(\delta\) -estimate holds true. $$f(x)=\frac{x-1}{4 x^{2}+1}, \quad\left[-\frac{3}{4}, 1\right], \quad a=\frac{1}{2}$$

Is there a value of \(b\) that will make $$g(x)=\left\\{\begin{array}{ll} x+b, & x < 0 \\ \cos x, & x \geq 0 \end{array}\right.$$ continuous at \(x=0 ?\) Differentiable at \(x=0 ?\) Give reasons for your answers.

Use a CAS to estimate the magnitude of the error in using the linearization in place of the function over a specified interval \(I\). Perform the following steps: a. Plot the function \(f\) over \(I\) b. Find the linearization \(L\) of the function at the point \(a\) c. Plot \(f\) and \(L\) together on a single graph. d. Plot the absolute error \(|f(x)-L(x)|\) over \(I\) and find its maximum value. e. From your graph in part (d), estimate as large a \(\delta>0\) as you can, satisfying \(|x-a|<\delta \quad \Rightarrow \quad|f(x)-L(x)|<\epsilon\) for \(\epsilon=0.5,0.1,\) and \(0.01 .\) Then check graphically to see if your \(\delta\) -estimate holds true. $$f(x)=x^{3}+x^{2}-2 x, \quad[-1,2], \quad a=1$$

What happens to the derivatives of \(\sin x\) and \(\cos x\) if \(x\) is measured in degrees instead of radians? To find out, take the following steps. a. With your graphing calculator or computer grapher in degree mode, graph $$f(h)=\frac{\sin h}{h}$$ and estimate \(\lim _{h \rightarrow 0} f(h) .\) Compare your estimate with \(\pi / 180 .\) Is there any reason to believe the limit should be \(\pi / 180 ?\) b. With your grapher still in degree mode, estimate $$\lim _{h \rightarrow 0} \frac{\cos h-1}{h}$$ c. Now go back to the derivation of the formula for the derivative of \(\sin x\) in the text and carry out the steps of the derivation using degree-mode limits. What formula do you obtain for the derivative? d. Work through the derivation of the formula for the derivative of \(\cos x\) using degree-mode limits. What formula do you obtain for the derivative? e. The disadvantages of the degree-mode formulas become apparent as you start taking derivatives of higher order. Try it. What are the second and third degree-mode derivatives of \(\sin x\) and \(\cos x ?\)

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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