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Chapter 3: Problem 25

A function \(h(x)\) is defined in terms of a differentiable \(f(x) .\) Find an expression for \(h^{\prime}(x).\) $$h(x)=\frac{f\left(x^{2}\right)}{x}$$

Short Answer

Expert verified
The expression for \( h'(x) \) is \( 2 f'(x^2) - \frac{f(x^2)}{x^2} \).

Step by step solution

01

Identify the function to be derived

The given function is \( h(x) = \frac{f(x^2)}{x} \). This is a composition of functions involving a quotient.
02

Apply the Quotient Rule

The Quotient Rule states that \( \left( \frac{u}{v} \right)' = \frac{u'v - uv'}{v^2} \) where \( u = f(x^2) \) and \( v = x \).
03

Compute the derivative of the numerator

To find \( u' \), use the Chain Rule because \( u = f(x^2) \). Let \( g(x) = x^2 \), thus \( u = f(g(x)) \). Then, \( u' = f'(g(x)) \cdot g'(x) = f'(x^2) \cdot 2x \).
04

Compute the derivative of the denominator

The derivative of \( v = x \) is \( v' = 1 \).
05

Substitute into the Quotient Rule

Substitute \( u, u' \), and \( v \) into the Quotient Rule formula to get: \[ h'(x) = \frac{f'(x^2) \cdot 2x \cdot x - f(x^2) \cdot 1}{x^2} \].
06

Simplify the expression

Simplify the numerator: \( 2x^2 f'(x^2) - f(x^2) \). Thus, \[ h'(x) = \frac{2x^2 f'(x^2) - f(x^2)}{x^2} = 2 f'(x^2) - \frac{f(x^2)}{x^2} \].

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Key Concepts

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

Understanding the Quotient Rule
When dealing with the differentiation of a function given as a quotient, we use the Quotient Rule. The Quotient Rule is a formula that helps find the derivative of a function that is the ratio of two differentiable functions. Mathematically, if you have a function \(\frac{u}{v}\), where both \(u\) and \(v\) are functions of \(x\), the Quotient Rule states: \[ \frac{d}{dx}\bigg(\frac{u}{v}\bigg) = \frac{u'v - uv'}{v^2} \]. Here鈥檚 a simple breakdown of the Law:
  • \(\frac{u}{v}\) is your original function.
  • \(u\) is the numerator, and \(v\) is the denominator.
  • You'll need to find \(u'\) and \(v'\), which are the derivatives of \(u\) and \(v\) respectively.
  • Substitute these into the Quotient Rule formula.
It is crucial to apply this rule carefully, especially when the numerator or the denominator is a composite function.
Mastering the Chain Rule
The Chain Rule is an essential tool in calculus for differentiating composite functions. A composite function is a function made by combining two functions, where one function is applied to the results of the other. For example, if \(y = f(g(x))\), we use the Chain Rule to find its derivative. The Chain Rule formula is: \[ \frac{dy}{dx} = f'(g(x)) \times g'(x) \]. Here鈥檚 how to approach it:
  • Identify the outer function \(f\) and the inner function \(g\).
  • Differentiate the outer function \(f\) with respect to the inner function \(g(x)\).
  • Multiply this by the derivative of the inner function \(g'(x)\).
In our exercise, we noticed that \(u = f(x^2)\), requiring us to use the Chain Rule to differentiate it. By letting \(g(x) = x^2\) and \(u = f(g(x))\), we find that \(u' = f'(x^2) \times 2x\). This result is then utilized in further calculations.
Exploring Function Composition
Function composition involves creating a new function by applying one function to the results of another. If you have two functions \(f(x)\) and \(g(x)\), the composite function is written as \(f(g(x))\). To evaluate this, you first apply \(g(x)\) and then apply \(f\) to the result. Understanding function composition is vital when dealing with complex differentiation problems. Here are some key points:
  • Composition often creates nested functions requiring careful differentiation.
  • Identify the inner and outer functions to apply the Chain Rule correctly.
In our example, \(h(x) = \frac{f(x^2)}{x}\), \(f(x^2)\) represents a composite function where \(x^2\) is inside \(f\). Proper handling of such compositions simplifies the application of differentiation rules like the Quotient Rule and Chain Rule.

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Most popular questions from this chapter

Use the fact that at the beginning of 1998, the population of the United States was 268,924,000 people and growing at the rate of 1,856,000 people per year. At the beginning of 1998, the annual consumption of ice cream in the United States was 12,582,000 pints and growing at the rate of 212 million pints per year. At what rate was the annual per capita consumption of ice cream increasing at that time? (Hint: [annual per capita consumption] \(=\frac{\text { [annual consumption] }}{\text { [population sized }}\).)

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Suppose that the price \(p\) (in dollars) and the weekly sales \(x\) (in thousands of units) of a certain commodity satisfy the demand equation $$2 p^{3}+x^{2}=4500.$$ Determine the rate at which sales are changing at a time when \(x=50, p=10,\) and the price is falling at the rate of \(\$ .50\) per week.

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