/*! 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 25 \(7 - 46\) Find the derivative o... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

\(7 - 46\) Find the derivative of the function. $$F ( z ) = \sqrt { \frac { z - 1 } { z + 1 } }$$

Short Answer

Expert verified
The derivative is \( \frac{1}{(z+1)\sqrt{(z-1)(z+1)}} \)."

Step by step solution

01

Rewrite the Function for Differentiation

We start by rewriting the given function \( F(z) = \sqrt{\frac{z - 1}{z + 1}} \) as \( F(z) = \left(\frac{z - 1}{z + 1}\right)^{1/2} \). This form allows us to apply the power rule and chain rule effectively.
02

Apply the Chain Rule

To differentiate \( F(z) = \left( \frac{z - 1}{z + 1} \right)^{1/2} \), apply the chain rule. Let \( u = \frac{z - 1}{z + 1} \), so \( F(z) = u^{1/2} \). The derivative is \( \frac{dF}{dz} = \frac{1}{2} u^{-1/2} \cdot \frac{du}{dz} \).
03

Find the Derivative of the Inner Function

Differentiate the inner function \( u = \frac{z-1}{z+1} \) using the quotient rule: \( \frac{du}{dz} = \frac{(z+1)(1) - (z-1)(1)}{(z+1)^2} = \frac{2}{(z+1)^2} \).
04

Combine the Derivatives

Substitute \( \frac{du}{dz} \) back into \( \frac{dF}{dz} = \frac{1}{2} \cdot u^{-1/2} \cdot \frac{2}{(z+1)^2} \). Simplify to get \( \frac{dF}{dz} = \frac{1}{\sqrt{u}} \cdot \frac{1}{(z+1)^2} \).
05

Substitute Back and Simplify

Recall that \( u = \frac{z-1}{z+1} \). Thus, \( \sqrt{u} = \sqrt{\frac{z-1}{z+1}} \). Substitute back to get \[ \frac{dF}{dz} = \frac{1}{\sqrt{\frac{z-1}{z+1}}} \cdot \frac{1}{(z+1)^2} = \frac{\sqrt{z+1}}{\sqrt{z-1}} \cdot \frac{1}{(z+1)^2} \]. Simplifying further gives the final derivative: \[ \frac{dF}{dz} = \frac{1}{(z+1)\sqrt{(z-1)(z+1)}} \].

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.

Chain Rule
The chain rule is a fundamental tool in calculus used for differentiating composite functions. A composite function is a function that arises when one function is inside another function. When you have a composite function like this, the chain rule helps us find the derivative by breaking down the process into manageable steps.
To use the chain rule, imagine we have a function \( F(x) = g(h(x)) \). The derivative of \( F \) with respect to \( x \) is found by taking the derivative of \( g \) with respect to \( h \), and then multiplying it by the derivative of \( h \) with respect to \( x \). In mathematical terms, this is expressed as:
  • \( \frac{d}{dx} F(x) = \frac{dg}{dh} \cdot \frac{dh}{dx} \)
In our problem, the function \( F(z) = \left(\frac{z-1}{z+1}\right)^{1/2} \) required the chain rule because it involved an outer function \( u^{1/2} \) where \( u = \frac{z-1}{z+1} \). We needed to find the derivative of the whole while focusing on \( u \) first then \( z \).The chain rule is powerful because it simplifies the process of differentiating complex functions into simpler tasks. Just remember to tackle the function from the outside in, using the derivative of each layer as it peels away.
Power Rule
The power rule is one of the simplest and most widely used rules in calculus for differentiation. It makes finding the derivative of a polynomial or a function with powers straightforward. The power rule states that if you have a function of the form \( x^n \), where \( n \) is any real number, then the derivative is:
  • \( \frac{d}{dx} x^n = n \cdot x^{n-1} \)
In our exercise, the function \( F(z) = \left(\frac{z-1}{z+1}\right)^{1/2} \) was transformed into a form more suitable for differentiation by using the power rule. The term \( \left(\frac{z-1}{z+1}\right)^{1/2} \) can be treated as \( u^{1/2} \) after letting \( u = \frac{z-1}{z+1} \).
Applying the power rule to differentiate this, we get:
  • \( \frac{d}{du} u^{1/2} = \frac{1}{2} u^{-1/2} \)
This result was then multiplied by the derivative of \( u \) with respect to \( z \), according to the chain rule. In problems involving powers, always look for an opportunity to use the power rule as it will simplify finding derivatives.
It's a straightforward technique that saves a lot of time, especially when dealing with polynomial terms or expressions with fractional exponents.
Quotient Rule
The quotient rule is essential for differentiating functions that are expressed as one function divided by another, essentially a fraction. When tackling a function in the form of \( g(x)/h(x) \), the quotient rule provides a formula for finding its derivative.The quotient rule is given by:
  • \( \frac{d}{dx} \frac{g(x)}{h(x)} = \frac{g'(x) h(x) - g(x) h'(x)}{(h(x))^2} \)
In our specific exercise, the inner function \( u = \frac{z - 1}{z + 1} \) required the quotient rule for its differentiation. By applying the quotient rule, we find:
  • \( \frac{du}{dz} = \frac{(z+1)(1) - (z-1)(1)}{(z+1)^2} = \frac{2}{(z+1)^2} \)
Note that the numerator results from multiplying the derivative of the top expression by the bottom expression and subtracting the product of the top expression by the derivative of the bottom.
The quotient rule is perfect for differentiating fractional functions because it systematically handles both the numerator and denominator, ensuring all components are considered. It's crucial to keep track of each part of the fraction and remember to square the bottom function in the denominator, ensuring accuracy in your calculations.

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

\(19-22\) Compute \(\Delta y\) and dy for the given values of \(x\) and \(d x=\Delta x\) . Then sketch a diagram like Figure 5 showing the line segments with lengths dx, dy, and \(\Delta y\) . \(y=e^{x}, x=0, \quad \Delta x=0.5\)

In a fish farm, a population of fish is introduced into a pond and harvested regularly. A model for the rate of change of the fish population is given by the equation $$\frac{\mathrm{dP}}{\mathrm{dt}}=\mathrm{r}_{0}\left(1-\frac{\mathrm{P}(\mathrm{t})}{\mathrm{P}_{\mathrm{c}}}\right) \mathrm{P}(\mathrm{t})-\beta \mathrm{P}(\mathrm{t})$$ where \(r_{0}\) is the birth rate of the fish, \(P_{c}\) is the maximum population that the pond can sustain (called the carrying capacity), and \(\beta\) is the percentage of the population that is harvested. (a) What value of dP/dt corresponds to a stable population? (b) If the pond can sustain \(10,000\) fish, the birth rate is 5\(\%\) . and the harvesting rate is \(4 \%,\) find the stable population level. (c) What happens if \(\beta\) is raised to 5\(\% ?\)

Two sides of a triangle are 4 \(\mathrm{m}\) and 5 \(\mathrm{m}\) in length and the angle between them is increasing at a rate of 0.06 \(\mathrm{rad} / \mathrm{s} .\) Find the rate at which the area of the triangle is increasing when the angle between the sides of fixed length is \(\pi / 3\) .

Find the derivative. Simplify where possible. $$y=\operatorname{coth}^{-1} \sqrt{x^{2}+1}$$

In the study of ecosystems, predator-prey models are often used to study the interaction between species. Consider populations of tundra wolves, given by \(\mathrm{W}(\mathrm{t}),\) and caribou, given by \(\mathrm{C}(\mathrm{t}),\) in northern Canada. The interaction has been modeled by the equations $$\frac{\mathrm{dC}}{\mathrm{dt}}=\mathrm{aC}-\mathrm{bCW} \quad \frac{\mathrm{dW}}{\mathrm{dt}}=-\mathrm{cW}+\mathrm{dCW}$$ (a) What values of \(\mathrm{dC} / \mathrm{dt}\) and dW/dt correspond to stable populations? (b) How would the statement "The caribou go extinct" be represented mathematically? (c) Suppose that a \(=0.05, \mathrm{b}=0.001, \mathrm{c}=0.05,\) and \(\mathrm{d}=0.0001 .\) Find all population pairs \((\mathrm{C}, \mathrm{W})\) that lead to stable populations. According to this model, is it possible for the two species to live in balance or will one or both species become extinct?

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.