/*! 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 22 Find the derivative of the follo... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 3: Problem 22

Find the derivative of the following functions. $$f(t)=6 \sqrt{t}-4 t^{3}+9$$

Short Answer

Expert verified
Answer: The derivative of the function is \(f'(t) = 3t^{-\frac{1}{2}} - 12t^2\).

Step by step solution

01

Identify the terms in the function

In this function, we have three terms: 1. \(6\sqrt{t}\) 2. \(-4t^3\) 3. \(9\) We will find the derivatives of each term separately and then combine them to find the derivative of the entire function.
02

Apply power rule to the first term

The first term is \(6\sqrt{t}\), which can be written as \(6t^{\frac{1}{2}}\). To find the derivative, apply the power rule, which states that \(\frac{d}{dt}(t^n) = nt^{n-1}\). Here: $$\frac{d}{dt}(6t^{\frac{1}{2}}) = 6 \cdot \frac{1}{2}t^{\frac{1}{2}-1} = 3t^{-\frac{1}{2}}$$
03

Apply power rule to the second term

The second term is \(-4t^3\). Applying the power rule, we get: $$\frac{d}{dt}(-4t^3) = -12t^2$$
04

Find the derivative of the third term

The third term is \(9\). The derivative of a constant is always 0. So the derivative of \(9\) is just \(0\).
05

Combine the derivatives of each term

Now, we can add the derivatives of each term to find the derivative of the entire function: $$\frac{d}{dt}(f(t)) = \frac{d}{dt}(6\sqrt{t} - 4t^3 + 9) = 3t^{-\frac{1}{2}} - 12t^2 + 0 = 3t^{-\frac{1}{2}} - 12t^2$$ So the derivative of the given function, \(f(t)\), is: $$f'(t) = 3t^{-\frac{1}{2}} - 12t^2$$

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
Understanding the power rule is essential when it comes to finding derivatives. The power rule is a helpful tool that simplifies the process of differentiation for functions of the form \(t^n\). According to the rule, if you want to find the derivative of \(t^n\), you simply bring down the exponent \(n\) as a multiplier and then decrease the exponent by one. This can be represented as:
\[\frac{d}{dt}(t^n) = nt^{n-1}\]
In the context of our problem, the power rule was applied to the terms \(6t^{\frac{1}{2}}\) and \(-4t^3\).
  • For \(6t^{\frac{1}{2}}\), the derivative becomes \(3t^{-\frac{1}{2}}\).
  • For \(-4t^3\), it becomes \(-12t^2\).
The power rule makes it straightforward to compute derivatives, especially when dealing with polynomial terms.
Constant Function
A constant function is one that always takes the same value, no matter what the input is. In mathematical terms, if \(f(t) = c\) where \(c\) is a constant, the derivative of this function is 0. This is because a constant function's graph is a horizontal line, indicating no change in the value as \(t\) changes.
In the given exercise, we see this concept in the term \(9\). Here, \(9\) is a constant, and its derivative is simply \(0\). This aligns perfectly with the rule that states the rate of change of a constant is always zero, since the value doesn't change regardless of any increase or decrease in \(t\). Recognizing constant functions and their derivatives can save you a lot of time as you tackle calculus problems involving complex functions.
Functions
Functions are a fundamental concept in calculus and mathematics in general. They are essentially relations between input values and outputs, often written as \(f(t)\). Each input \(t\) into the function has one corresponding output. In calculus, we often want to know how these outputs change as we modify the inputs, which is why finding derivatives is so crucial.
Functions come in various forms, such as linear, quadratic, polynomial, and more complex expressions. In our problem, \(f(t) = 6\sqrt{t} - 4t^3 + 9\) represents a function with three distinct terms.
  • \(6\sqrt{t}\) is a term that involves a square root, which needed to be rewritten as \(6t^{\frac{1}{2}}\) to make differentiation easier.
  • \(-4t^3\) is a polynomial term, and its derivative follows the familiar pattern thanks to the power rule.
  • \(9\) is a straightforward constant function.
Understanding the properties and types of functions will help you apply the correct differentiation rules effectively.

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

Work carefully Proceed with caution when using implicit differentiation to find points at which a curve has a specified slope. For the following curves, find the points on the curve (if they exist) at which the tangent line is horizontal or vertical. Once you have found possible points, make sure they actually lie on the curve. Confirm your results with a graph. $$x^{2}\left(3 y^{2}-2 y^{3}\right)=4$$

General logarithmic and exponential derivatives Compute the following derivatives. Use logarithmic differentiation where appropriate. $$\frac{d}{d x}\left(x^{\left(x^{10}\right)}\right)$$

Logistic growth Scientists often use the logistic growth function \(P(t)=\frac{P_{0} K}{P_{0}+\left(K-P_{0}\right) e^{-r_{d}}}\) to model population growth, where \(P_{0}\) is the initial population at time \(t=0, K\) is the carrying capacity, and \(r_{0}\) is the base growth rate. The carrying capacity is a theoretical upper bound on the total population that the surrounding environment can support. The figure shows the sigmoid (S-shaped) curve associated with a typical logistic model. World population (part 1 ) The population of the world reached 6 billion in \(1999(t=0)\). Assume Earth's carrying capacity is 15 billion and the base growth rate is \(r_{0}=0.025\) per year. a. Write a logistic growth function for the world's population (in billions) and graph your equation on the interval \(0 \leq t \leq 200\) using a graphing utility. b. What will the population be in the year 2020? When will it reach 12 billion?

Vertical tangent lines a. Determine the points where the curve \(x+y^{2}-y=1\) has a vertical tangent line (see Exercise 53 ). b. Does the curve have any horizontal tangent lines? Explain.

Let \(C(x)\) represent the cost of producing \(x\) items and \(p(x)\) be the sale price per item if \(x\) items are sold. The profit \(P(x)\) of selling x items is \(P(x)=x p(x)-C(x)\) (revenue minus costs). The average profit per item when \(x\) items are sold is \(P(x) / x\) and the marginal profit is dP/dx. The marginal profit approximates the profit obtained by selling one more item given that \(x\) items have already been sold. Consider the following cost functions \(C\) and price functions \(p\). a. Find the profit function \(P\). b. Find the average profit function and marginal profit function. c. Find the average profit and marginal profit if \(x=a\) units are sold. d. Interpret the meaning of the values obtained in part \((c)\). $$C(x)=-0.02 x^{2}+50 x+100, p(x)=100, a=500$$
See all solutions

Recommended explanations on Math Textbooks

Probability and Statistics

Read Explanation

Logic and Functions

Read Explanation

Pure Maths

Read Explanation

Statistics

Read Explanation

Geometry

Read Explanation

Decision Maths

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.