Problem 64 - Calculus and its applications

Question

Let \(f(u)=\frac{u+1}{u-1}\) and \(g(x)=u=\sqrt{x}\) Find \((f \circ g)^{\prime}(4)\)

Step-by-Step Solution

Answer

The derivative \\( (f \circ g)'(4) = \frac{-1}{2} \\).

Step 1: Understanding Composition

In this problem, we need to find the derivative of the composite function \(f \circ g\), where \(f(u)=\frac{u+1}{u-1}\) and \(g(x)=\sqrt{x}\). The composition \(f \circ g\) represents \(f(g(x))\).

Step 2: Express the Composite Function

First, substitute \(g(x)\) into \(f(u)\). This gives \((f \circ g)(x) = f(\sqrt{x}) = \frac{\sqrt{x} + 1}{\sqrt{x} - 1}\). This is the expression for the composite function.

Step 3: Differentiate the Composite Function

Use the quotient rule to differentiate \( (f \circ g)(x) = \frac{\sqrt{x} + 1}{\sqrt{x} - 1} \). The quotient rule states that if \(h(x) = \frac{n(x)}{d(x)}\), then \(h'(x) = \frac{n'(x)d(x) - n(x)d'(x)}{[d(x)]^2}\).Let \ n(x) = \sqrt{x} + 1 \ and \ d(x) = \sqrt{x} - 1\.First, find the derivatives:\ n'(x) = \frac{1}{2\sqrt{x}} \ \ d'(x) = \frac{1}{2\sqrt{x}} \ Substitute these into the quotient rule:\ (f \circ g)'(x) = \frac{(\frac{1}{2\sqrt{x}})(\sqrt{x} - 1) - (\sqrt{x} + 1)(\frac{1}{2\sqrt{x}})}{(\sqrt{x}-1)^2} \.

Step 4: Simplify the Derivative Expression

Simplify the numerator of \( (f \circ g)'(x)\):\ ((\sqrt{x} - 1) - (\sqrt{x} + 1))\frac{1}{2\sqrt{x}} = \frac{-2}{2\sqrt{x}} = \frac{-1}{\sqrt{x}} \.So, the expression becomes \((f \circ g)'(x) = \frac{-1}{\sqrt{x}(\sqrt{x}-1)^2}\).

Step 5: Evaluate the Derivative at x = 4

Substitute \(x = 4\) into the derivative expression:\(\sqrt{4} = 2\), so:\( (f \circ g)'(4) = \frac{-1}{2(2-1)^2} = \frac{-1}{2(1)^2} = \frac{-1}{2}\).

Key Concepts

Quotient Rule

The quotient rule is used to find the derivative of a quotient, which is a fraction where one function is divided by another. Let’s break it down simply with the case where you have two functions, say

• numerator function: \( n(x) \)
• denominator function: \( d(x) \)

When you have a function \( h(x) = \frac{n(x)}{d(x)} \), and you need to find the derivative, the quotient rule comes into play. The rule can be remembered with the formula:\[h'(x) = \frac{n'(x)\cdot d(x) - n(x)\cdot d'(x)}{[d(x)]^2}\]What this means is that you first find the derivative of the numerator and multiply it by the denominator. Then, you subtract the numerator multiplied by the derivative of the denominator. Importantly, you divide the entire result by the square of the denominator function.In the exercise, we applied the quotient rule to differentiate \((f \circ g)(x) = \frac{\sqrt{x} + 1}{\sqrt{x} - 1}\). This required identifying

• \(n(x) = \sqrt{x} + 1\)
• \(d(x) = \sqrt{x} - 1\)

and calculating their derivatives, then applying the quotient rule formula to get the final derivative form. This is how the operations work out efficiently!

Chain Rule

The chain rule is a fundamental technique for finding the derivative of composite functions. When you have a function nested inside another, such as in the form of \(f(g(x))\), you use the chain rule to differentiate. Here’s the basic idea: if you have an outer function \( f \) and an inner function \( g \), the derivative of the composite function \((f \circ g)(x)\) is:\[(f \circ g)'(x) = f'(g(x)) \cdot g'(x)\]This means you first take the derivative of the outer function evaluated at the inner function (\(g(x)\)), then multiply it by the derivative of the inner function itself. In our given problem, the function \(g(x) = \sqrt{x}\) serves as the inner function, while \(f(u) = \frac{u+1}{u-1}\) is the outer function. The chain rule subtly underlies how these composed functions interplay through differentiation, even though we utilized the quotient rule explicitly after substituting \(g\) into \(f\). Applying the chain rule ensures the entire setup of differentiating composite functions adheres to correct mathematical principles.

Composite Functions

Composite functions are developed when one function is applied to the result of another function, denoted as \(f(g(x))\) or \((f \circ g)(x)\). It’s like double layering functions: you feed \(x\) into \(g\), and then take the output of \(g\) as the input for \(f\). In the exercise, we constructed the composite function by substituting \(g(x) = \sqrt{x}\) into \(f(u) = \frac{u+1}{u-1}\), to yield:\[(f \circ g)(x) = \frac{\sqrt{x} + 1}{\sqrt{x} - 1}\]Creating composite functions can sometimes make solving problems more complex, as each function might add its own challenges, especially when differentiation is involved. Understanding the interaction between the inner and outer functions is key to breaking down the composite for any mathematical analysis.Remember, carefully substitute the inner into the outer, ensuring each step aligns correctly to avoid mistakes. This becomes particularly important for calculus operations like differentiation, where each component needs accurate handling to ensure precise results.