Problem 79 - University Calculus: Early Transcendentals

Question

In Exercises \(79-84,\) find the value of \((f \circ g)^{\prime}\) at the given value of \(x\). $$f(u)=u^{5}+1, u=g(x)=\sqrt{x}, \quad x=1$$

Step-by-Step Solution

Answer

\((f \circ g)'(1) = \frac{5}{2}\).

Step 1: Chain Rule Formula

To find the derivative of the composition of two functions, use the chain rule: \((f \circ g)'(x) = f'(g(x)) \cdot g'(x)\). This requires calculating the derivatives \(f'(u)\) and \(g'(x)\).

Step 2: Differentiate f(u)

Differentiate \(f(u) = u^5 + 1\) with respect to \(u\). The derivative is \(f'(u) = 5u^4\).

Step 3: Differentiate g(x)

Differentiate \(g(x) = \sqrt{x}\), which can be rewritten as \(x^{1/2}\). The derivative is \(g'(x) = \frac{1}{2}x^{-1/2} = \frac{1}{2\sqrt{x}}\).

Step 4: Evaluate g(1)

Calculate \(g(1)\) which is \(\sqrt{1} = 1\).

Step 5: Evaluate f'(g(1))

Substitute \(g(1) = 1\) into \(f'(u)\). This gives \(f'(1) = 5(1)^4 = 5\).

Step 6: Evaluate g'(1)

Substitute \(x = 1\) into \(g'(x)\). This gives \(g'(1) = \frac{1}{2\sqrt{1}} = \frac{1}{2}\).

Step 7: Calculate Composite Derivative

Now multiply \(f'(g(1))\) by \(g'(1)\) to find \((f \circ g)'(1)\): \((f \circ g)'(1) = 5 \cdot \frac{1}{2} = \frac{5}{2}\).

Key Concepts

Composite Functions

Composite functions involve taking two or more functions and combining them to form a new function. They are written in the form

• \((f \circ g)(x) = f(g(x))\)

This means that you first apply the function \(g(x)\), and then apply the function \(f\) to the result. In simpler terms, it is like performing an action (\(g\)) and then another action (\(f\)) on the result of the first action.
Composite functions can be tricky because they require understanding not just each individual function but also how they work together. This is crucial when trying to manipulate or differentiate these functions.
Remembering the importance of order in composite functions is key. The first function inside the parentheses \(g(x)\) is applied first, and its result then becomes the input for the outer function \(f(u)\). Understanding this layered approach helps in visualizing and solving problems involving composite functions.

Derivative Calculations

Derivative calculations involve finding the rate at which a function is changing at any point. The chain rule is particularly useful when dealing with composite functions, allowing us to differentiate them effectively.The chain rule states that if you have two functions such as \(f\) and \(g\), then the derivative of their composition \((f \circ g)\) is given by:

• \((f \circ g)'(x) = f'(g(x)) \cdot g'(x)\)

This means you need to

• First, differentiate the outer function \(f\) with respect to its variable \(u\)
• Then, differentiate the inner function \(g\) with respect to \(x\)

For example, in the given problem, we have \(f(u) = u^5 + 1\) and \(g(x) = \sqrt{x}\). When you find \(f'(u)\), it simplifies to \(5u^4\), and \(g'(x)\) simplifies to \(\frac{1}{2\sqrt{x}}\).
These derivatives must be evaluated at the specific value, which in this case is \(x=1\). The results are used in the chain rule equation to find the final derivative of the composite function: \((f \circ g)'(1) = \frac{5}{2}\). This process highlights how systematic and organized derivative calculations can lead to an accurate outcome.

Function Composition

Function composition refers to the operation of applying one function to the results of another. It allows for the simplification and reuse of functions.
The key aspect is defining a pathway from an input \(x\) through a sequence of functions. Consider the inner function \(g(x)\), which processes \(x\), and then the outcome is transformed further by the outer function \(f(u)\).
This task involves embedding one function within another, which can streamline complex operations. The composition of functions is not just computationally useful but also conceptually significant in connecting different mathematical ideas.

• Illustration: In our example, \(g(x) = \sqrt{x}\) is computed first, and the output becomes the input for \(f(u) = u^5 + 1\).

Function composition serves various mathematical purposes, including advanced calculus operations like finding limits and derivatives. Understanding how to navigate through function compositions is essential for mastering more complicated mathematical concepts. This approach enhances problem-solving skills by providing a powerful tool for working with an array of function-related challenges.