Chapter 2: Problem 49
In Exercises 39–52, find the derivative of the function. $$ f(x)=\sqrt{x}-6 \sqrt[3]{x} $$
Short Answer
Step by step solution
Key Concepts
These are the key concepts you need to understand to accurately answer the question.
/*! 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}
Learning Materials
Features
Discover
Chapter 2: Problem 49
In Exercises 39–52, find the derivative of the function. $$ f(x)=\sqrt{x}-6 \sqrt[3]{x} $$
These are the key concepts you need to understand to accurately answer the question.
All the tools & learning materials you need for study success - in one app.
Get started for free
Determining Differentiability In Exercises \(85-88\) , find the derivatives from the left and from the right at \(x=1\) (if they exist). Is the function differentiable at \(x=1 ?\) $$ f(x)=\sqrt{1-x^{2}} $$
Determining Differentiability In Exercises \(85-88\) , find the derivatives from the left and from the right at \(x=1\) (if they exist). Is the function differentiable at \(x=1 ?\) $$ f(x)=|x-1| $$
Horizontal Tangent Line In Exercises \(73-76\) , determine the point(s) at which the graph of the function has a horizontal tangent line. $$ f(x)=\frac{x^{2}}{x-1} $$
Determining Differentiability In Exercises \(85-88\) , find the derivatives from the left and from the right at \(x=1\) (if they exist). Is the function differentiable at \(x=1 ?\) $$ f(x)=\left\\{\begin{array}{ll}{(x-1)^{3},} & {x \leq 1} \\ {(x-1)^{2},} & {x>1}\end{array}\right. $$
Moving Point In Exercises \(5-8,\) a point is moving along the graph of the given function at the rate \(d x / d t .\) Find \(d y / d t\) for the given values of \(x .\) $$ \begin{array}{l}{y=\cos x ; \frac{d x}{d t}=4 \text { centimeters per second }} \\ {\begin{array}{llll}{\text { (a) } x=\frac{\pi}{6}} & {\text { (b) } x=\frac{\pi}{4}} & {\text { (c) } x=\frac{\pi}{3}}\end{array}}\end{array} $$
What do you think about this solution?
We value your feedback to improve our textbook solutions.