Problem 51 - Calculus Single Variable

Question

Let \(c=4, F(x)=1+3 x, G(x)=\sqrt{x}\) and \(H(x)=x /(x+5) .\) Calculate the requested derivative. $$ (F \circ G \circ H)^{\prime}(c) $$

Step-by-Step Solution

Answer

The derivative \((F \circ G \circ H)'(4)\) is \(\frac{5}{36}\).

Step 1: Understand the Problem

The task is to compute the derivative of a composition of functions, specifically \((F \circ G \circ H)(x)\), and evaluate it at \(c = 4\). This involves applying the chain rule for derivatives.

Step 2: Define the Composite Function

The composite function \((F \circ G \circ H)(x)\) can be expressed as \(F(G(H(x)))\). Our functions are defined as follows: \(F(x) = 1 + 3x\), \(G(x) = \sqrt{x}\), and \(H(x) = \frac{x}{x+5}\). Substitute these into the composite function.

Step 3: Evaluate H

First, evaluate \(H(x)\): \[H(x) = \frac{x}{x+5}\] Specially, at \(x = c = 4\), it becomes \(H(4) = \frac{4}{4+5} = \frac{4}{9}\).

Step 4: Evaluate G

Next, use \(H(4)\) to evaluate \(G\): \[G(H(4)) = G\left(\frac{4}{9}\right) = \sqrt{\frac{4}{9}} = \frac{2}{3}\].

Step 5: Evaluate F

Now use \(G(H(4))\) to evaluate \(F\): \[F(G(H(x))) = F\left(\frac{2}{3}\right) = 1 + 3 \times \frac{2}{3} = 1 + 2 = 3\].

Step 6: Differentiate H(x)

Differentiate \(H(x) = \frac{x}{x+5}\) using the quotient rule: \[H'(x) = \frac{(x+5)\cdot1 - x\cdot1}{(x+5)^2} = \frac{5}{(x+5)^2}\]. At \(x = 4\), \[H'(4) = \frac{5}{(4+5)^2} = \frac{5}{81}\].

Step 7: Differentiate G(x)

Differentiate \(G(x) = \sqrt{x}\) with respect to \(x\): \[G'(x) = \frac{1}{2\sqrt{x}}\]. At \(x = \frac{4}{9}\), \[G'(\frac{4}{9}) = \frac{1}{2\sqrt{\frac{4}{9}}} = \frac{3}{4} = \frac{3}{4}\].

Step 8: Differentiate F(x)

Differentiate \(F(x) = 1 + 3x\): \[F'(x) = 3\].

Step 9: Apply the Chain Rule

The derivative of the composite function \((F \circ G \circ H)'(x)\) using the Chain Rule is:\[(F \circ G \circ H)'(c) = F'(G(H(c))) \cdot G'(H(c)) \cdot H'(c)\]. Substitute the values: \[(F \circ G \circ H)'(4) = 3 \cdot \frac{3}{4} \cdot \frac{5}{81} = \frac{15}{108} = \frac{5}{36}\].

Key Concepts

Derivative of Composite Functions

When we talk about the derivative of composite functions, we're diving into the realm of calculus where functions are nested within each other. Imagine it as a set of Russian dolls, where each doll is inside a larger one. Similarly, in mathematical terms, the derivative of a composite function involves differentiating a function inside another function. This is often handled using the chain rule, which is a powerful technique that allows us to differentiate composite functions step-by-step.

Here's how the process works:

• First, identify each function in the composition, such as in our exercise where we have functions \( F(x) \), \( G(x) \), and \( H(x) \).
• Then, differentiate each function individually. This means computing the derivatives \( F'(x) \), \( G'(x) \), and \( H'(x) \).
• Finally, apply the chain rule which involves multiplying these derivatives as shown: \( F'(G(H(x))) \, G'(H(x)) \, H'(x) \).

By following these steps, we simplify what initially seems complex. Each derivative builds upon the previous one, providing us with a comprehensive expression for the derivative of the entire composite function. When evaluated at a specific point, like \( x = 4 \) in our example, the result gives us the rate of change of the composite function at that point.

Quotient Rule

The quotient rule is another essential tool in the calculus toolkit, especially when dealing with ratios of functions. Whenever you have a function that is a division of two other functions, the quotient rule helps you find its derivative. This rule states that if you have a function \( H(x) = \frac{u(x)}{v(x)} \), then its derivative is given by:

• \( H'(x) = \frac{v(x)u'(x) - u(x)v'(x)}{(v(x))^2} \)

This equation might look intimidating at first glance, but breaking it down can help:

• Consider each part of the fraction as its own function: the numerator \( u(x) \) and the denominator \( v(x) \).
• Differentiate each of these functions separately to get \( u'(x) \) and \( v'(x) \).
• Plug these derivatives into the quotient rule formula to find the derivative of the overall function.

In our exercise, \( H(x) = \frac{x}{x+5} \), and using the quotient rule, we found \( H'(x) = \frac{5}{(x+5)^2} \). This gives us a clear understanding of how the function \( H(x) \) changes at any point \( x \), especially highlighted when substituting \( x = 4 \) for final evaluation.

Function Composition

Function composition is a way to build more complex functions by combining simpler ones. It is a fundamental concept in mathematics, allowing us to explore how one process can influence another. Imagine each function as an operation or a machine. One machine processes the input, and the resulting output becomes the input for the next machine.

In our particular exercise, we are dealing with three functions \( F(x) \), \( G(x) \), and \( H(x) \), each performing a specific role:

• \( H(x) = \frac{x}{x+5} \) acts as the initial operator by modifying \( x \) into a fraction.
• This output is then fed into \( G(x) = \sqrt{x} \), which takes the square root of that result.
• Finally, \( F(x) = 1 + 3x \) scales and shifts the result to give the final output.

By composing these functions together, we can carry out a sequence of operations through a single input and produce a final result. It showcases the power of function composition in creating pathways from input to output, ultimately illustrating the deep interconnectedness of mathematical functions. This interconnected pathway is elegantly unraveled through differentiation when applying the chain rule to composite functions.