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91Ó°ÊÓ

Find \(f^{\prime}(x)\). $$ f(x)=\frac{5}{x}-x^{2 / 3} $$

Short Answer

Expert verified
\( f^{\prime}(x) = -\frac{5}{x^2} - \frac{2}{3}x^{-1/3} \).

Step by step solution

01

Rewrite the Function

The given function is \( f(x) = \frac{5}{x} - x^{2/3} \). First, rewrite the terms using exponents: \( f(x) = 5x^{-1} - x^{2/3} \).
02

Differentiate Each Term

Apply the power rule to differentiate each term separately. For \( 5x^{-1} \), the derivative is \( -5x^{-2} \). For \( x^{2/3} \), the derivative is \( \frac{2}{3}x^{-1/3} \).
03

Combine the Derivatives

Combine the derivatives from the previous step. Thus, \( f^{\prime}(x) = -5x^{-2} - \frac{2}{3}x^{-1/3} \).
04

Simplify the Expression

Combine the terms to form a single expression for the derivative: \( f^{\prime}(x) = -\frac{5}{x^2} - \frac{2}{3}x^{-1/3} \). This is the simplified form of the derivative.

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Key Concepts

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

Understanding Calculus
Calculus is a branch of mathematics that deals with change. It allows us to understand how functions behave and how they change over different values of input variables. Calculus consists of two main branches:
  • Differential calculus, which deals with rates of change and slopes of curves,
  • Integral calculus, which is concerned with accumulation of quantities and areas under curves.
Differential calculus is often used to find the derivative of a function. The derivative provides useful information about the rate of change of the function with respect to its variables. This is particularly important in fields like physics, engineering, and economics, where understanding how something changes is crucial to predicting behavior.
By recognizing how different types of functions can change, calculus enables us to solve complex problems ranging from computing velocity to optimizing production costs.
  • Learners must become familiar with different function types and operations, such as polynomials, exponential, and logarithmic functions.
  • Conceptual understanding is key: grasp what a derivative represents before diving into calculations.
Applying the Power Rule
The power rule is one of the most commonly used techniques in differentiation. It provides a simple formula to find the derivative of functions that are polynomials or any function that can be expressed as a power of a variable.
The power rule states that for any function of the form:
\( f(x) = x^n \),
the derivative is:
\( f'(x) = nx^{n-1} \).
In the original exercise, the power rule was used to find the derivative of each component of the function. Here is how it applies:
  • For \( 5x^{-1} \), applying the power rule yields \( -5x^{-2} \), because the original exponent was -1, so you multiply by it and then subtract one from the exponent.
  • For \( x^{2/3} \), the derivative using the power rule is \( \frac{2}{3}x^{-1/3} \), similar steps are performed as above by bringing down the exponent and reducing it by one.
Understanding and effectively applying the power rule is critical to performing calculus problems with efficiency. It simplifies finding derivatives and reduces the time needed to work through problems.
Basics of Differentiation
Differentiation is a fundamental concept in calculus. It involves computing the derivative of a function, which represents the function's rate of change. Mastering differentiation is essential for analyzing and interpreting the behavior of various functions over time.
The process of differentiation involves applying specific rules and formulas to evaluate the slope or rate of change of functions. The derivative tells us how a function changes as its input changes, which is critical for many applications in science and engineering.
  • The first rule often learned is the power rule, but other rules include the product rule, quotient rule, and chain rule.
  • Each rule is useful for different types of functions and compositions.
  • In practical terms, finding a derivative is much like finding the slope of a tangent line to a point on a function's graph.
In the exercise given, converting fractions into negative exponents was a step to easily apply the power rule. This small preparation step is typical when working with expressions in calculus that involve differentiating polynomial types and similar functions.

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Most popular questions from this chapter

Find the interval(s) for which \(f^{\prime}(x)\) is positive. Find the points on the graph of $$ y=x^{4}-\frac{4}{3} x^{2}-4 $$

The circumference \(C,\) in centimeters, of a healing wound is approximated by $$ C(r)=6.28 r $$ where \(r\) is the wound's radius, in centimeters. a) Find the rate of change of the circumference with respect to the radius. b) Find \(C^{\prime}(4)\) c) Explain the meaning of your answer to part (b).

Graph the function \(f\) given by $$ f(x)=\left\\{\begin{array}{ll} -3, & \text { for } x=-2 \\ x^{2}, & \text { for } x \neq-2 \end{array}\right. $$ Use GRAPH and TRACE to find each of the following limits. When necessary, state that the limit does not exist. a) \(\lim _{x \rightarrow-2^{+}} f(x)\) b) \(\lim _{x \rightarrow-2^{-}} f(x)\) c) \(\lim _{x \rightarrow-2} f(x)\) d) \(\lim _{x \rightarrow 2^{+}} f(x)\) e) \(\lim _{x \rightarrow 2^{-}} f(x)\) f) Does \(\lim _{x \rightarrow-2} f(x)=f(-2)\) ? g) Does \(\lim _{x \rightarrow 2} f(x)=f(2)\) ?

It has been shown that the home range, in hectares, of a carnivorous mammal weighing \(w\) grams can be approximated by $$ H(w)=0.11 w^{1.36} $$ (Source: Based on information in Emlen, J. M., Ecology: An Evolutionary Approach, p. \(216,\) Reading, MA: Addison-Wesley, 1973 ; and Harestad, A. S., and Bunnel, F. L., "Home Range and Body Weight-A Reevaluation," Ecology, Vol. 60, No. 2, pp. 405-418.) a) Find the average rate at which a carnivorous mammal's home range increases as the animal's weight grows from \(500 \mathrm{~g}\) to \(700 \mathrm{~g}\). b) Find \(\frac{H(300)-H(200)}{300-200}\). What does this rate represent?

Let \(f\) and \(g\) be differentiable over an open interval containing \(x=a\). If $$ \lim _{x \rightarrow a}\left(\frac{f(x)}{g(x)}\right)=\frac{0}{0} \quad \text { or } \quad \lim _{x \rightarrow a}\left(\frac{f(x)}{g(x)}\right)=\frac{\pm \infty}{\pm \infty} $$ and if \(\lim _{x \rightarrow a}\left(\frac{f^{\prime}(x)}{g^{\prime}(x)}\right)\) exists, then $$ \lim _{x \rightarrow a}\left(\frac{f(x)}{g(x)}\right)=\lim _{x \rightarrow a}\left(\frac{f^{\prime}(x)}{g^{\prime}(x)}\right) $$ The forms \(0 / 0\) and \(\pm \infty / \pm \infty\) are said to be indeterminate. In such cases, the limit may exist, and l'Hôpital's Rule offers a way to find the limit using differentiation. For example, in Example 1 of Section \(1.1,\) we showed that $$ \lim _{x \rightarrow 1}\left(\frac{x^{2}-1}{x-1}\right)=2 $$ Since, for \(x=1,\) we have \(\left(x^{2}-1\right)(x-1)=0 / 0,\) we differentiate the numerator and denominator separately, and reevaluate the limit: $$ \lim _{x \rightarrow 1}\left(\frac{x^{2}-1}{x-1}\right)=\lim _{x \rightarrow 1}\left(\frac{2 x}{1}\right)=2 $$ Use this method to find the following limits. Be sure to check that the initial substitution results in an indeterminate form. $$ \lim _{x \rightarrow \infty}\left(\frac{4 x^{2}+x-3}{2 x^{2}+1}\right) $$

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