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Chapter 3: Problem 22

\(3-26\) Differentiate. $$g(t)=\frac{t-\sqrt{t}}{t^{1 / 3}}$$

Short Answer

Expert verified
The derivative is \(g'(t) = \frac{2}{3}t^{-1/3} - \frac{1}{6}t^{-5/6}\).

Step by step solution

01

Simplify the Expression

First, we need to simplify the given function. The expression is given by \[g(t) = \frac{t - \sqrt{t}}{t^{1/3}}\]Rewrite the terms individually:1. \(\frac{t}{t^{1/3}} = t^{1 - 1/3} = t^{2/3}\)2. \(\frac{\sqrt{t}}{t^{1/3}} = \frac{t^{1/2}}{t^{1/3}} = t^{1/2 - 1/3} = t^{3/6 - 2/6} = t^{1/6}\)Thus, the function simplifies to:\[g(t) = t^{2/3} - t^{1/6}\]
02

Differentiate the Simplified Function

Apply the power rule of differentiation, which states that if \(f(t) = t^n\), then \(f'(t) = n \cdot t^{n-1}\).1. Differentiate \(t^{2/3}\): \[\frac{d}{dt}(t^{2/3}) = \frac{2}{3}t^{-1/3}\]2. Differentiate \(t^{1/6}\): \[\frac{d}{dt}(t^{1/6}) = \frac{1}{6}t^{-5/6}\]So, the derivative of \(g(t)\) is:\[g'(t) = \frac{2}{3}t^{-1/3} - \frac{1}{6}t^{-5/6}\]
03

Present the Final Answer

The derivative of the original function \(g(t) = \frac{t - \sqrt{t}}{t^{1/3}}\) is:\[g'(t) = \frac{2}{3}t^{-1/3} - \frac{1}{6}t^{-5/6}\]

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Key Concepts

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

Power Rule
The Power Rule is a fundamental principle in calculus used to find the derivative of a function of the form \( f(t) = t^n \). The rule states that the derivative of \( t^n \) with respect to \( t \) is \( n \cdot t^{n-1} \).
This simple rule is incredibly powerful and forms the basis of much of differential calculus.
  • For example, when applying the power rule to \( t^{2/3} \), we multiply \( 2/3 \) by \( t \) and subtract 1 from the exponent, leading to \( \frac{2}{3}t^{-1/3} \).
  • Similarly, for the term \( t^{1/6} \), we get a derivative of \( \frac{1}{6}t^{-5/6} \).
These transformations are straightforward once you grasp the key concept: multiply by the exponent and decrease the exponent by one. This repeatable step makes the power rule a go-to technique for differentiating powers of \( t \).
Simplification of Expressions
Before differentiating a function, it's often useful or necessary to simplify the expression. Simplification can make differentiation easier or even possible if the function is complex.
In the exercise, the function \( g(t) = \frac{t - \sqrt{t}}{t^{1/3}} \) is simplified by individually rewriting the terms:
  • \( \frac{t}{t^{1/3}} = t^{2/3} \) involves subtracting the exponents \( 1 - 1/3 \).
  • \( \frac{\sqrt{t}}{t^{1/3}} = t^{1/6} \) is simplified by computing \( 1/2 - 1/3 = 1/6 \).
Simplifying expressions helps reveal the underlying structure of a function, making the differentiation process straightforward. Keeping a function in its simplest form can avoid complications during calculation.
Derivative Calculation
The process of calculating derivatives involves applying rules like the power rule to an expression that represents a function. In the given exercise, we start with the simplified function \( g(t) = t^{2/3} - t^{1/6} \). By separately finding the derivatives of each term and then combining them, we arrive at the overall derivative.
  • For \( t^{2/3} \), the derivative is \( \frac{2}{3}t^{-1/3} \).
  • For \( t^{1/6} \), the derivative is \( \frac{1}{6}t^{-5/6} \).
The final derivative is then expressed as \( g'(t) = \frac{2}{3}t^{-1/3} - \frac{1}{6}t^{-5/6} \). Calculating derivatives requires attention to detail but employing established rules simplifies the process. Understanding the rules minimizes errors and builds confidence in handling more complex derivatives in the future.

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Most popular questions from this chapter

Find \(\frac{d^{9}}{d x^{9}}\left(x^{8} \ln x\right)\)

The frequency of vibrations of a vibrating violin string is given by $$\mathrm{f}=\frac{1}{2 \mathrm{L}} \sqrt{\frac{\mathrm{T}}{\rho}}$$ where \(L\) is the length of the string, T is its tension, and \(\rho\) is its linear density. [See Chapter 11 in D. E. Hall, Musical Acoustics, 3 \(\mathrm{d}\) ed. (Pacific Grove, CA: Brooks/Cole, \(2002 ) . ]\) (a) Find the rate of change of the frequency with respect to (i) the length (when T and \(\rho\) are constant), (ii) the tension (when L and \(\rho\) are constant), and (iii) the linear density (when L and T are constant). (b) The pitch of a note (how high or low the note sounds) is determined by the frequency f. (The higher the frequency, the higher the pitch.) Use the signs of the derivatives in part (a) to determine what happens to the pitch of a note (i) when the effective length of a string is decreased by placing a finger on the string so a shorter portion of the string vibrates, (ii) when the tension is increased by turning a tuning peg. (iii) when the linear density is increased by switching to another string.

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