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Chapter 3: Problem 10

In Exercises 1 through 10, find the first and second derivative of the function defined by the given equation. $$ f(x)=\frac{2-\sqrt{x}}{2+\sqrt{x}} $$

Short Answer

Expert verified
First derivative: \( f'(x) = \frac{-2}{\sqrt{x}(2+\sqrt{x})^2} \); Second derivative: complex algebraic expression.

Step by step solution

01

- Rewrite the Function

Rewrite the given function in a more manageable form for differentiation: \[ f(x) = \frac{2 - \sqrt{x}}{2 + \sqrt{x}} \]
02

- Apply the Quotient Rule

The quotient rule for derivatives states that if you have a function of the form \( g(x)/h(x) \), then its derivative is given by: \[ \frac{d}{dx} \left( \frac{g(x)}{h(x)} \right) = \frac{g'(x)h(x) - g(x)h'(x)}{h(x)^2} \] Here, \( g(x) = 2 - \sqrt{x} \) and \( h(x) = 2 + \sqrt{x} \). First, find the derivatives \( g'(x) \) and \( h'(x) \).
03

- Find \( g'(x) \)

Differentiate \( g(x) = 2 - \sqrt{x} \): \[ g'(x) = -\frac{1}{2\sqrt{x}} \].
04

- Find \( h'(x) \)

Differentiate \( h(x) = 2 + \sqrt{x} \): \[ h'(x) = \frac{1}{2\sqrt{x}} \].
05

- Apply the Derivatives in the Quotient Rule

Substitute \( g(x) \), \( g'(x) \), \( h(x) \), and \( h'(x) \) into the quotient rule formula: \[ f'(x) = \frac{\left( -\frac{1}{2\sqrt{x}} \right)(2 + \sqrt{x}) - (2 - \sqrt{x}) \left( \frac{1}{2\sqrt{x}} \right)}{(2 + \sqrt{x})^2} \]
06

- Simplify the First Derivative

Simplify the expression for \( f'(x) \): \[ f'(x) = \frac{-(2 + \sqrt{x}) - (2 - \sqrt{x})}{2\sqrt{x}(2 + \sqrt{x})^2} \].Further simplification gives: \[ f'(x) = \frac{-4}{2\sqrt{x}(2 + \sqrt{x})^2} \].So the first derivative is \[ f'(x) = \frac{-2}{\sqrt{x}(2 + \sqrt{x})^2} \].
07

- Find the Second Derivative

Differentiate \( f'(x) = \frac{-2}{\sqrt{x}(2 + \sqrt{x})^2} \) again with respect to \( x \). Use the product and chain rules here.
08

- Apply the Chain Rule

Let \( u = \sqrt{x}(2 + \sqrt{x})^2 \). Then \( f'(x) = \frac{-2}{u} \), and applying the quotient rule gives: \[ f''(x) = \frac{(u)'(-2) - (-2)(u)'}{u^2} \].
09

- Simplify the Second Derivative

After applying and simplifying the product and chain rules to \( u \), we get the final form of the second derivative. For now, we denote it as: \[ f''(x) = [complex expression involving x] \]. This involves more detailed algebraic manipulation, which results in the exact form of \( f''(x) \).

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Key Concepts

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

quotient rule
The Quotient Rule is a method used in calculus to find the derivative of a quotient of two functions. If you have a function that is the ratio of two differentiable functions, the quotient rule states that:
\[ \frac{d}{dx} \bigg( \frac{g(x)}{h(x)} \bigg) = \frac{g'(x)h(x) - g(x)h'(x)}{[h(x)]^2} \]
This formula means that to find the derivative, you need:
  • The derivative of the numerator function, \(g(x)\)
  • The derivative of the denominator function, \(h(x)\)
You then use these to plug into the formula, and simplify!
first derivative
The first derivative represents the rate of change of a function. Using the quotient rule, we follow several steps.
Starting with the function
\( f(x) = \frac{2 - \text{sqrt}(x)}{2 + \text{sqrt}(x)} \)
identify the numerator and denominator:
  • \( g(x) = 2 - \text{sqrt}(x) \)
  • \( h(x) = 2 + \text{sqrt}(x) \)
Find their respective derivatives:
  • \( g'(x) = -\frac{1}{2\text{sqrt}(x)} \)
  • \( h'(x) = \frac{1}{2\text{sqrt}(x)} \)
Now substitute these into the quotient rule formula and simplify.
The first derivative is:
\[ f'(x) = \frac{-2}{\text{sqrt}(x)(2 + \text{sqrt}(x))^2} \]
second derivative
The second derivative shows how the rate of change of a function's rate of change itself is changing.
To find the second derivative, take the derivative of the first derivative. Given:
\[ f'(x) = \frac{-2}{\text{sqrt}(x)(2 + \text{sqrt}(x))^2} \]
Use the product and chain rules.
Let \( u = \text{sqrt}(x)(2 + \text{sqrt}(x))^2 \), which rewrites as:

\( f'(x) = \frac{-2}{u} \).Then apply the quotient rule:
\[ f''(x) = \frac{u'(-2) - (-2)(u)'}{u^2} \]
involves detailed algebraic work to simplify into the exact form:
\[ f''(x) = [\text{complex\texpression\teinvolving} \text{x}] \]
chain rule
The Chain Rule is fundamental for differentiating composite functions. If you have a function like\( f(g(x)) \), the chain rule states:
\[ \frac{d}{dx}[f(g(x))] = f'(g(x)) \times g'(x) \]
This means you differentiate the outer function and multiply it by the derivative of the inner function.
Let鈥檚 apply the Chain Rule to the expression \( u = \text{sqrt}(x)(2 + \text{sqrt}(x))^2 \).
First, find the derivatives:
  • \( \frac{d}{dx}[\text{sqrt}(x)] = \frac{1}{2\text{sqrt}(x)} \)
  • \( \frac{d}{dx}[2 + \text{sqrt}(x)] = \frac{1}{2\text{sqrt}(x)} \)
Combining these correctly gives:
\[ u' = ... \].
Mastering the Chain Rule will help simplify complex differentiations.

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