/*! 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 28 Find the derivative. It may be t... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Find the derivative. It may be to your advantage to simplify before differentiating. Assume \(a, b, c,\) and \(k\) are constants. $$h(w)=w \arcsin w$$

Short Answer

Expert verified
The derivative of \( h(w) = w \arcsin w \) is \( h'(w) = \arcsin w + \frac{w}{\sqrt{1-w^2}} \).

Step by step solution

01

Simplify the Expression

The given function is \( h(w) = w \arcsin w \). While this expression might not benefit significantly from algebraic simplification before differentiation, it's important to note that it is in a product form which suggests we might use the product rule.
02

Apply the Product Rule

The product rule for differentiation states that \( (uv)' = u'v + uv' \) where \( u \) and \( v \) are functions of \( w \). In this case, let \( u = w \) and \( v = \arcsin w \). We need to differentiate both \( u \) and \( v \) individually.
03

Differentiate \( u = w \)

The derivative of \( u = w \) with respect to \( w \) is \( u' = 1 \).
04

Differentiate \( v = \arcsin w \)

The derivative of \( v = \arcsin w \) with respect to \( w \) is \( v' = \frac{1}{\sqrt{1-w^2}} \).
05

Combine Results Using the Product Rule

Using the product rule, we get:\[(u'v + uv') = (1)\arcsin w + w\left(\frac{1}{\sqrt{1-w^2}}\right)\]Thus, the derivative of \( h(w) = w \arcsin w \) is:\[h'(w) = \arcsin w + \frac{w}{\sqrt{1-w^2}}\]
06

Final Answer for the Derivative

Thus, the derivative of the given function is:\[h'(w) = \arcsin w + \frac{w}{\sqrt{1-w^2}}\]

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 91Ó°ÊÓ!

Key Concepts

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

Product Rule
The product rule is a valuable tool in calculus for differentiating functions that are a product of two or more simpler functions. If you have two functions, say \( u(w) \) and \( v(w) \), both depending on \( w \), the derivative of their product \( u(w) v(w) \) with respect to \( w \) is given by:
  • \( (uv)' = u'v + uv' \)
This means you take the derivative of the first function \( u \) while leaving \( v \) undifferentiated, and then add the product of the undifferentiated \( u \) with the derivative of \( v \).
In the exercise, \( h(w) = w \arcsin w \), we let \( u = w \) and \( v = \arcsin w \), where \( u' = 1 \) and \( v' = \frac{1}{\sqrt{1-w^2}} \).

So, by the product rule, the derivative \( h'(w) \) becomes \( u'v + uv' = \arcsin w + \frac{w}{\sqrt{1-w^2}} \).
This step-by-step approach helps to ensure all parts of the function are appropriately differentiated.
Arcsin Function
The arcsin function is the inverse of the sine function, usually denoted as \( \arcsin w \). It means given a sine value \( w \), the function returns the angle \( \theta \) within the range \([-\frac{\pi}{2}, \frac{\pi}{2}]\).
When differentiating the arcsin function with respect to \( w \), this results in:
  • \( \frac{d}{dw}(\arcsin w) = \frac{1}{\sqrt{1-w^2}} \)
This derivative formula emerges because of the trigonometry properties and limits applied to \( \arcsin \).
It is important to remember that \( -1 \leq w \leq 1 \), since outside this domain the arcsin function is not defined in the real numbers.
The derivative \( \frac{1}{\sqrt{1-w^2}} \) shows how sensitive the arcsin function is to changes in \( w \). Python and other programming languages allow direct calculations for such functions.
Simplification
Simplification means reducing expressions to a form that is easier to work with or interpret. In calculus, before differentiating, it can sometimes help to simplify the expression.
Although simplification might not drastically alter the way we differentiate \( h(w) = w \arcsin w \), recognizing it as a product of \( w \) and \( \arcsin w \) directs us to the correct usage of the product rule.
Steps for simplification often include:
  • Factoring expressions.
  • Cancelling common terms.
  • Rewriting functions using known identities or relationships.
For \( h(w) \), nothing simplifies further without changing the form of the equation we set out to differentiate.
It's a gentle reminder that sometimes simplifying doesn't mean making the expression shorter; it means making it more comprehensible for the processes we intend to perform on it.
Differentiation
The process of differentiation involves finding how a function changes at any point, essentially measuring the rate of change. It's fundamental in calculus for analyzing the behavior of functions.
Differentiating \( h(w) = w \arcsin w \) involves applying the product rule, as these are two distinct functions multiplying each other.
Differentiation steps include:
  • Identifying which rules apply, like the product rule in this case.
  • Computing individual derivatives: \( \frac{d}{dw}w = 1 \) and \( \frac{d}{dw}(\arcsin w) = \frac{1}{\sqrt{1-w^2}} \).
  • Combining these using the product rule to get the overall derivative.
This process is crucial in solving real-world problems where changes and trends need quantifying, like in physics for speed calculations, or in economics for cost analysis.
Understanding the roles of each function in the product and applying the differentiation rules appropriately unveils the slope or rate of change for \( h(w) \), offering insights into its behavior over its domain.

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

Are the statements true or false for a function \(f\) whose domain is all real numbers? If a statement is true, explain how you know. If a statement is false, give a counterexample. If \(f^{\prime}(x) \leq 1\) for all \(x\) and \(f(0)=0,\) then \(f(x) \leq x\) for all \(x\)

Find a formula for the error \(E(x)\) in the tangent line approximation to the function near \(x=a\). Using a table of values for \(E(x) /(x-a)\) near \(x=a\), find a value of \(k\) such that \(E(x) /(x-a) \approx k(x-a) .\) Check that, approximately, \(k=f^{\prime \prime}(a) / 2\) and that \(E(x) \approx\left(f^{\prime \prime}(a) / 2\right)(x-a)^{2}\). $$f(x)=\cos x, a=0$$

Find the equations of the tangent lines to the graph of \(f(x)=\sin x\) at \(x=0\) and at \(x=\pi / 3 .\) Use each tangent line to approximate \(\sin (\pi / 6) .\) Would you expect these results to be equally accurate, since they are taken equally far away from \(x=\pi / 6\) but on opposite sides? If the accuracy is different, can you account for the difference?

(a) Using a calculator or computer, sketch the graph of \(y=2 e^{x}+5 e^{-x}\) for \(-3 \leq x \leq 3,0 \leq y \leq 20\) Observe that it looks like the graph of \(y=\cosh x\) Approximately where is its minimum? (b) Show algebraically that \(y=2 e^{x}+5 e^{-x}\) can be written in the form \(y=A \cosh (x-c) .\) Calculate the values of \(A\) and \(c .\) Explain what this tells you about the graph in part (a).

Give an example of: A function \(f\) that is differentiable on the interval (0,2) but does not satisfy the conclusion of the Mean Value Theorem on the interval [0,2]

See all solutions

Recommended explanations on Math Textbooks

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.