/*! 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 35 Suppose that \(f^{\prime}(x)=\fr... [FREE SOLUTION] | 91影视

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Chapter 4: Problem 35

Suppose that \(f^{\prime}(x)=\frac{1}{x}\). Find the following: (a) \(\frac{d}{d x} f\left(x^{2}+3\right)\) (b) \(\frac{d}{d x} f(\sqrt{x-1})\)

Short Answer

Expert verified
(a) \( \frac{2x}{x^2 + 3} \); (b) \( \frac{1}{2(x-1)} \)."

Step by step solution

01

Understand the Problem

We are given that \( f'(x) = \frac{1}{x} \). We need to find the derivatives \( \frac{d}{d x} f\left(x^{2}+3\right) \) and \( \frac{d}{d x} f(\sqrt{x-1}) \). First, we recognize that these require the application of the chain rule in differentiation.
02

Differentiate Part (a) Using Chain Rule

For (a), we need to find \( \frac{d}{dx} f(x^2 + 3) \). By the chain rule, if \( g(x) = x^2 + 3 \), then \( \frac{d}{dx} f(g(x)) = f'(g(x)) \cdot g'(x) \). Here, \( f'(g(x)) = \frac{1}{g(x)} \) and \( g'(x) = 2x \). Thus, the derivative is \[ \frac{d}{dx} f(x^2 + 3) = \frac{1}{x^2 + 3} \cdot 2x = \frac{2x}{x^2 + 3}. \]
03

Differentiate Part (b) Using Chain Rule

For (b), we need \( \frac{d}{dx} f(\sqrt{x-1}) \). Let \( h(x) = \sqrt{x-1} = (x-1)^{1/2} \). By the chain rule, \( \frac{d}{dx} f(h(x)) = f'(h(x)) \cdot h'(x) \). Here, \( f'(h(x)) = \frac{1}{h(x)} \) and \( h'(x) = \frac{1}{2}(x-1)^{-1/2} \). Thus, the derivative is \[ \frac{d}{dx} f(\sqrt{x-1}) = \frac{1}{\sqrt{x-1}} \cdot \frac{1}{2\sqrt{x-1}} = \frac{1}{2(x-1)}. \]

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Key Concepts

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

Chain Rule
The chain rule is a powerful technique in calculus. It's used to differentiate composite functions. When you have a function within another function, like \( f(g(x)) \), the chain rule becomes handy. This rule states that to find the derivative of \( f(g(x)) \), you multiply the derivative of the outer function \( f' \) by the derivative of the inner function \( g' \).

Here's the formula expressed in more detail:
  • Let \( y = f(u) \), where \( u = g(x) \).
  • Then \( \frac{dy}{dx} = \frac{dy}{du} \times \frac{du}{dx} \).
To understand better, think of peeling an onion. You have to peel each layer one by one. Similarly, in the chain rule, you differentiate each layer starting from the outermost to the innermost. Each "layer" is a separate function that you work with.
Derivatives
Derivatives tell us the rate at which a function is changing. It鈥檚 like watching the speedometer of a car to understand how fast it鈥檚 going at any moment.

To calculate a derivative, you typically use rules based on the form of the function. The common rules include:
  • Power Rule: If \( f(x) = x^n \), then \( f'(x) = nx^{n-1} \).
  • Product Rule: For functions \( u(x) \) and \( v(x) \), \( (uv)' = u'v + uv' \).
  • Quotient Rule: For \( u(x) \) and \( v(x) \), \( (\frac{u}{v})' = \frac{u'v - uv'}{v^2} \).
  • Chain Rule: Already explained above.
Derivatives are fundamental in calculus because they help in understanding instantaneous change. For example, with \( f'(x) = \frac{1}{x} \), it shows how the function \( f \) changes as \( x \) changes. The exercises we solved use these principles to find the derivatives of complex functions.
Function Notation
Function notation is a way to express mathematical relationships. Think of it like a machine: you input a number, and the machine (function) transforms it into another number.

In function notation, \( f(x) \) indicates a function named \( f \) with \( x \) as the input variable.
  • For example, if \( f(x) = x^2 + 3 \), then \( f(x) \) tells you what to do with \( x \), specifically to square \( x \) and add 3.
  • The notation \( f(g(x)) \) means you first apply \( g(x) \) to \( x \), then apply \( f(x) \) to the result.
  • This is critical when using the chain rule because you need to understand which function operates inside another.
Understanding how to use function notation is crucial in calculus. It provides clarity and structure when working with complex math problems, especially when differentiating functions.

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Most popular questions from this chapter

Find the derivative with respect to the independent variable. $$ f(x)=2 \cos \left(x^{3}-3 x\right) $$

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Consider the chemical reaction $$ \mathrm{A}+\mathrm{B} \longrightarrow \mathrm{AB} $$ If \(x(t)\) denotes the concentration of \(\mathrm{AB}\) at time \(t\), and \(k\) is the rate constant for of the reaction, explain why: $$ \frac{d x}{d t}=k(a-x)(b-x) $$ where \(k\) is a positive constant and \(a\) and \(b\) denote the concentrations of \(\mathrm{A}\) and \(\mathrm{B}\), respectively, at time 0 .

Wing Beat Frequency in Hummingbirds Altshuler and Dudley (2003) found that hummingbirds' wing beat frequency, \(f\), decreases with body mass, \(m\), according to $$ f=40-\frac{8}{5} m $$ where \(f\) is measured in beats per second and \(m\) in grams. Assume that the amount of thrust that a flying hummingbird can generate depends on its mass and wing beat frequency as follows $$ T=c f^{2} m^{4 / 3} $$ for some positive constant \(c\). (This equation is derived from the thrust mechanics of a moving wing.) (a) Equation (4.5) should only be used if \(m<25\). Why? (The largest hummingbirds have a mass of \(22 \mathrm{~g}\).) (b) Calculate \(d T / d m .\) (c) Show by plotting \(d T / d m\) that for larger hummingbirds, \(T\) decreases with \(m\). That is, show that \(d T / d m<0\), once \(m\) exceeds a critical threshold. (d) For a hummingbird to fly, its thrust must exceed its weight \(W=m g\) (where \(g\) is the acceleration due to gravity). Explain, using your answer to (d), why if the hummingbird's mass increases, then \(T\) will eventually be smaller than \(W\) (Hint: \(d W / d m=g\) is constant and positive), no matter what the value of \(c\) is.
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