/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 34 Find \(f^{\prime}(x)\). $$f(x)... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 1: Problem 34

Find \(f^{\prime}(x)\). $$f(x)=0.3 x^{1.2}$$

Short Answer

Expert verified
The derivative is \( f^{\rime}(x) = 0.36 \, x^{0.2} \).

Step by step solution

01

Understand the function

The given function is \( f(x) = 0.3 \, x^{1.2} \). We need to find its derivative.
02

Use the power rule for differentiation

The power rule for differentiation states that if \( f(x) = a \, x^{b} \), then \( f^{\rime}(x) = a \, b \, x^{b-1} \).
03

Differentiate the function

Apply the power rule to differentiate \( f(x) = 0.3 \, x^{1.2} \): \[ f^{\rime}(x) = 0.3 \, \times \, 1.2 \, \times \, x^{1.2 - 1} = 0.36 \, x^{0.2} \].

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

power rule
The power rule is a fundamental tool in differentiation. It simplifies taking the derivative of power functions. If you have a function in the form of \( f(x) = a \, x^{b} \), where \( a \) and \( b \) are constants, the power rule provides us with a straightforward way to find its derivative. The rule states:
  • Formula: \( f^{\rime}(x) = a \, b \, x^{b-1} \).

Let's break this down with an example. If \( f(x) = 0.3 \, x^{1.2} \), by applying the power rule, we multiply the exponent (1.2) by the coefficient (0.3) and then reduce the exponent by one:
  • \(0.3 \, \times \, 1.2 = 0.36\)
  • The new exponent is \(1.2 - 1 = 0.2\)
Therefore, the derivative is \( f^{\rime}(x) = 0.36 \, x^{0.2} \). The power rule helps make differentiation quick and easy.
derivative
A derivative represents the rate at which a function is changing at any given point. Think of it as the slope of the tangent line to the function's graph at a specific point. In mathematical terms, if you have a function \( f(x) \), the derivative, denoted as \( f^{\rime}(x) \) or \( \frac{df}{dx} \), shows how \( f(x) \) changes as \( x \) changes.
  • Basic Idea: For a small change in \( x \), the function changes by \( f^{\rime}(x) \, \times \, \text{small change in} \, x \).
  • Notation: Common symbols for derivatives include \( f^{\rime}(x) \), \( \frac{df}{dx} \) and \( Df(x) \).
Returning to our function \( f(x) = 0.3 \, x^{1.2} \), finding the derivative involved using the power rule. Thus we have \( f^{\rime}(x) = 0.36 \, x^{0.2} \). This tells us how the function \( f(x) \) changes with respect to \( x \), which is essential in many applications across physics, engineering, economics, and beyond.
calculus
Calculus is a branch of mathematics focused on studying changes. It's divided into two main parts: differentiation and integration. Differentiation, which we explored with the power rule and derivatives, deals with finding the rate of change of functions.
  • Differentiation: This part of calculus helps us understand how a function's output changes as its input changes. It involves concepts like the power rule to find derivatives.
  • Integration: This is the reverse process of differentiation. It deals with finding the total accumulation of quantities, such as areas under curves.
In our example, by differentiating \( f(x) = 0.3 \, x^{1.2} \), we used fundamental calculus principles to find \( f^{\rime}(x) = 0.36 \, x^{0.2} \). Calculus tools allow us to analyze everything from the motion of planets to optimizing business profits, making it an invaluable field of study.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

If \(f(x)\) is a function, then \((f \circ f)(x)=f(f(x))\) is the composition of \(f\) with itself. This is called an iterated function, and the composition can be repeated many times. For example, \((f \circ f \circ f)(x)=f(f(f(x))) .\) Iterated functions are very useful in many areas, including finance (compound interest is \(a\) simple case) and the sciences (in weather forecasting, for example). For the each function, use the Chain Rule to find the derivative.. If \(f(x)=x+\sqrt{x},\) find \(\frac{d}{d x}[(f \circ f)(x)]\).

Use the Chain Rule to differentiate each function. You may need to apply the rule more than once. $$f(x)=\left(2 x^{5}+(4 x-5)^{2}\right)^{6}$$

Differentiate each function. $$f(x)=\frac{x^{-1}}{x+x^{-1}}$$

If \(f(x)\) is a function, then \((f \circ f)(x)=f(f(x))\) is the composition of \(f\) with itself. This is called an iterated function, and the composition can be repeated many times. For example, \((f \circ f \circ f)(x)=f(f(f(x))) .\) Iterated functions are very useful in many areas, including finance (compound interest is \(a\) simple case) and the sciences (in weather forecasting, for example). For the each function, use the Chain Rule to find the derivative. If \(f(x)=x^{2}+1,\) find \(\frac{d}{d x}[(f \circ f)(x)]\).

Graph \(f\) and \(f^{\prime}\) Then estimate points at which the tangent line to \(f\) is horizontal. If no such point exists, state that fact. $$f(x)=\frac{x^{3}-1}{x^{2}+1}$$
See all solutions

Recommended explanations on Math Textbooks

Discrete Mathematics

Read Explanation

Decision Maths

Read Explanation

Statistics

Read Explanation

Probability and Statistics

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Geometry

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.