Problem 26 - The Calculus with Analytic Geometry

Question

Given \(f(u)=u^{2}+5 u+5\) and \(g(x)=(x+1) /(x-1)\). Find the derivative of \(f \circ g\) in two ways: (a) by first finding \((f \circ g)(x)\) and then finding \((f \circ g)^{\prime}(x) ;\) (b) by using the chain rule.

Step-by-Step Solution

Answer

The derivative \((f \circ g)'(x)\) is \(\left(2 \cdot \frac{x+1}{x-1} + 5\right) \cdot \frac{-2}{(x-1)^2}\).

Step 1: Understand the functions

Given two functions are \(f(u) = u^2 + 5u + 5\) and \(g(x) = \frac{x+1}{x-1}\). The goal is to find the derivative of the composite function \(f \circ g\) using two methods.

Step 2: Find \((f \circ g)(x)\)

First, find the expression for \((f \circ g)(x)\), which means substituting \(g(x)\) into \(f(u)\).\[ (f \circ g)(x) = f(g(x)) = f\left(\frac{x+1}{x-1}\right) \]

Step 3: Substitute and simplify

Substitute \(g(x)\) into \(f(u)\),\[ f \left( \frac{x+1}{x-1} \right) = \left( \frac{x+1}{x-1} \right)^2 + 5 \cdot \left( \frac{x+1}{x-1} \right) + 5 \]

Step 4: Differentiate \((f \circ g)(x)\)

Differentiate \((f \circ g)(x)\) with respect to \(x\).Using the quotient rule and chain rule, we find:\((f \circ g)'(x)\).

Step 5: Use the chain rule

Use the chain rule to find the derivative of \(f \circ g\). The chain rule states: \( (f \circ g)'(x) = f'(g(x)) \cdot g'(x) \).Find \(f'(u)\) and \(g'(x)\).

Step 6: Differentiate \(f\) and \(g\)

Find the derivatives:\[ f'(u) = 2u + 5 \]\[ g'(x) = \frac{d}{dx} \left( \frac{x+1}{x-1} \right) = \frac{(x-1) - (x+1)}{(x-1)^2} = \frac{-2}{(x-1)^2} \]

Step 7: Apply the chain rule

Substitute back into the chain rule equation:\[(f \circ g)'(x) = f'(g(x)) \cdot g'(x)\]\[ = (2 \cdot \frac{x+1}{x-1} + 5) \cdot \frac{-2}{(x-1)^2}\]

Step 8: Simplify the expression

Combine and simplify:\[ (f \circ g)'(x) = \left(2 \cdot \frac{x+1}{x-1} + 5\right) \cdot \frac{-2}{(x-1)^2}\]

Key Concepts

Composite Functions

A composite function is formed when one function is applied to the result of another function. For two functions, say, f(u) and g(x), the composite function can be written as \( (f \circ g)(x) \). This means that you first apply the function g to x, and then apply the function f to the result of g(x).
In mathematical terms, \( (f \circ g)(x) = f(g(x)) \).
To understand this concept better, think of it as a machine where g(x) is the first machine that processes raw material (input x), and then sends the output to machine f(u), which then processes it further.

Chain Rule

The chain rule is a fundamental technique for differentiating composite functions. It states that the derivative of the composite function \( (f \circ g)(x) \) is the derivative of f with respect to u, evaluated at g(x), multiplied by the derivative of g with respect to x.
The formula for the chain rule is expressed as:
\[ (f \circ g)'(x) = f'(g(x)) \cdot g'(x) \]
This rule helps to break down complex differentiation tasks into simpler parts.
For example, if you need to differentiate \( f(g(x)) \), you find \( f'(u) \), then evaluate it at \( u = g(x) \), and finally multiply it by \( g'(x) \).

Quotient Rule

The quotient rule is used to differentiate functions that are divided by one another. If you have a function \( g(x) \) in the form of a quotient such as \[ h(x) = \frac{u(x)}{v(x)} \], then the quotient rule states:
\[ h'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{(v(x))^2} \]
In plain terms, you take the derivative of the numerator (top function), multiply it by the denominator (bottom function), subtract the product of the numerator and the derivative of the denominator, and then divide the whole result by the square of the denominator.

Function Composition

Function composition involves creating a new function by applying one function to the results of another. Essentially, this means taking two functions, such as f and g, and combining them into a composite function like \( f \circ g \).
For example, given \[ f(u) = u^2 + 5u + 5 \] and \[ g(x) = \frac{x+1}{x-1} \], you can create a composite function \[ (f \circ g)(x) \] by substituting \ g(x) \ into \ f(u) \:
\[ (f \circ g)(x) = f(g(x)) = f\left( \frac{x+1}{x-1} \right) = \left( \frac{x+1}{x-1} \right)^2 + 5 \cdot \left( \frac{x+1}{x-1} \right) + 5 \]
Function composition is a useful concept in various areas of calculus and helps in simplifying complex expressions.

Derivatives

Derivatives measure how a function changes as its input changes. In more technical terms, the derivative of a function at a particular point gives the slope of the tangent line to the function's graph at that point.
We denote the derivative of a function f(x) as \ f'(x) \ or \frac{df}{dx} \. It represents the rate of change of the function f with respect to the variable x.
When dealing with composite functions, derivatives can be computed using rules like the chain rule and the quotient rule.
For example, given \[ f(u) = u^2 + 5u + 5 \] and \[ g(x) = \frac{x+1}{x-1} \] and using the chain rule, we find:
\ f'(u) = 2u + 5 \
\ g'(x) = \frac{-2}{(x-1)^2} \
So, \[ (f \circ g)'(x) = f'(g(x)) \cdot g'(x) = \left( 2 \cdot \frac{x+1}{x-1} + 5 \right) \cdot \frac{-2}{(x-1)^2} \]
Understanding derivatives is essential in calculus for solving various problems involving rates of change, optimization, and more.