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

Short Answer

Expert verified
The derivative \(\frac{dv}{dt} = 1 + \frac{1}{t^2}\).

Step by step solution

01

Understand the Given Function

Identify the function we need to differentiate. Here, the function is given as \( v = t - \frac{1}{t} \). This means \(v\) is expressed in terms of the variable \(t\).
02

Apply Derivative Rules

To find the derivative \( \frac{dv}{dt} \), differentiate the function \( v = t - \frac{1}{t} \) with respect to \( t \). The function can be separated into two parts: \( t \) and \(-\frac{1}{t}\).
03

Differentiate Each Part Individually

Differentiate \( t \) with respect to \( t \). The derivative of \( t \) with respect to \( t \) is 1, because the derivative of \(x\) with respect to \(x\) is always 1. Now, differentiate \(-\frac{1}{t}\). This can be rewritten as \(-t^{-1}\).
04

Apply Power Rule to \(-t^{-1}\)

Use the power rule for derivatives, which states that the derivative of \( x^n \) with respect to \( x \) is \( nx^{n-1} \). For \(-t^{-1}\), apply this rule: the derivative is \(-(-1)t^{-2} = \frac{1}{t^2}\).
05

Combine the Derivatives

After differentiating each part, combine the results: \(\frac{d}{dt}(t - \frac{1}{t}) = 1 + \frac{1}{t^2}\). This is the derivative of the function \( v \) with respect to \( t \).

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Key Concepts

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

Differentiation Techniques
Differentiating a function involves finding its derivative, which is essentially the rate at which the function changes with respect to a variable. In calculus, there are several techniques to differentiate functions, based on the form of the function. One common approach is to identify and apply derivative rules, such as the power rule, product rule, quotient rule, and chain rule, among others. These techniques help simplify the process of finding derivatives, especially when dealing with complex expressions.

For our function, which is expressed as a simple sum and difference of terms, differentiation can be quite straightforward. Here, recognizing the form of each part of the function allows us to apply specific rules directly to find the derivative efficiently. Understanding which differentiation technique to apply is foundational to tackling more complex calculus problems.
Power Rule
The power rule is a fundamental tool in calculus for finding the derivative of expressions in which a variable is raised to a power. According to the power rule, if you have a function of the form \( x^n \), its derivative is \( nx^{n-1} \). This means you bring down the power as a coefficient and reduce the power by one.
  • For a simple function like \( t \), which is actually \( t^1 \), its derivative using the power rule is simply \( 1 \times t^{1-1} = 1 \).
  • In the case of \(-\frac{1}{t}\), we can rewrite this as \(-t^{-1}\). Using the power rule, the derivative becomes \( -1 \times t^{-2} = -\frac{1}{t^2} \).
Breaking down each term and applying the power rule simplifies finding the derivative, making it an indispensable technique for various differentiation problems.
Variable Substitution
Variable substitution is a strategy often used in calculus to simplify functions before differentiation. While our original function \( v = t - \frac{1}{t} \) doesn't necessarily need substitution due to its simplicity, variable substitution becomes crucial when dealing with functions that are complex or difficult to differentiate directly.

This technique involves replacing complex parts of a function with a new variable, making it easier to apply standard differentiation rules. Once the differentiation is done, you substitute back the original variables. Although not directly required in this exercise, developing a strong understanding of this method prepares you for more advanced calculus topics.
Chain Rule
The chain rule is another differentiation technique that helps in handling functions composed of multiple functions. It's widely used when you encounter a function inside another function—in other words, a composite function. The chain rule states that the derivative of a composite function \( f(g(x)) \) is \( f'(g(x)) \cdot g'(x) \).

While the given exercise didn't require the chain rule due to its straightforward form, understanding this concept is crucial for differentiating more complex expressions, especially those involving nested functions. Mastery of the chain rule expands your ability to solve intricate calculus problems with ease.

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Most popular questions from this chapter

Find the derivatives of the functions in Exercises \(19-40\) $$ f(\theta)=\left(\frac{\sin \theta}{1+\cos \theta}\right)^{2} $$

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Radians versus degrees: degree mode derivatives 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 \(\cos x ?\)

If \(r=\sin (f(t)), f(0)=\pi / 3,\) and \(f^{\prime}(0)=4,\) then what is \(d r / d t\) at \(t=0 ?\)

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