Problem 53 - Thomas Calculus

Question

In Exercises \(53-58,\) find the value of \((f \circ g)^{\prime}\) at the given value of \(x\) . $$ f(u)=u^{5}+1, \quad u=g(x)=\sqrt{x}, \quad x=1 $$

Step-by-Step Solution

Answer

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

Step 1: Understand the Function Composition

We are given two functions, \( f(u) = u^5 + 1 \) and \( g(x) = \sqrt{x} \). We are asked to find the derivative of their composition, \((f \circ g)\), at \( x = 1 \). In function composition, \((f \circ g)(x) = f(g(x))\).

Step 2: Express the Composite Function

Substitute \( g(x) = \sqrt{x} \) into \( f(u) \). This gives us:\[ (f \circ g)(x) = f(\sqrt{x}) = (\sqrt{x})^5 + 1 = x^{5/2} + 1 \]

Step 3: Differentiate the Composite Function

Differentiate \( (f \circ g)(x) = x^{5/2} + 1 \) with respect to \( x \). The derivative is:\[ \frac{d}{dx}[x^{5/2} + 1] = \frac{5}{2}x^{3/2} \]

Step 4: Evaluate the Derivative at x=1

Substitute \( x = 1 \) into the derivative:\[ \left. \frac{5}{2}x^{3/2} \right|_{x=1} = \frac{5}{2}(1)^{3/2} = \frac{5}{2}\]

Step 5: Summarize the Result

The derivative of the composite function \( (f \circ g) \) at \( x = 1 \) is \( \frac{5}{2} \).

Key Concepts

Function Composition

Function composition is a fundamental concept in calculus that involves combining two or more functions. When you compose two functions, such as \( f \) and \( g \), you are essentially plugging one function into another. This is represented as \( (f \circ g)(x) = f(g(x)) \).
This means that whatever output we have from \( g(x) \) becomes the input for \( f(u) \). In the given exercise, \( f(u) = u^5 + 1 \) and \( g(x) = \sqrt{x} \), so the composite function is \( (f \circ g)(x) = f(\sqrt{x}) = (\sqrt{x})^5 + 1 \).

• Think of function composition like using multiple filters on an image; the first filter changes the image and the second develops it further.
• In mathematical terms, if the function \( g \) modifies \( x \) by some method, then \( f \) further processes \( g(x) \).

Understanding the result of function composition is crucial when progressing to computing derivatives.

Derivative

The derivative of a function is a core concept in calculus that signifies how a function changes when the input changes. In simpler terms, it is the rate of change or the slope of the function at a certain point.
To find the derivative of a composite function like \( (f \circ g)(x) = x^{5/2} + 1 \), you apply the rules of differentiation to each part of the function. Here, we differentiate \( x^{5/2} \) to get \( \frac{5}{2}x^{3/2} \) since the derivative of \( x^n \) is \( nx^{n-1} \).
This portion of a calculus problem helps us understand how outputs vary with slight changes in inputs.

• Using the power rule is essential here, which states \( \frac{d}{dx} x^n = nx^{n-1} \).
• Always apply the differentiation rules according to the form of each term in the function.

This step is crucial before moving on to evaluate the derivative at a specific point.

Evaluation at a Point

Once you have the derivative of a function, it's often useful to evaluate that derivative at a specific point. This means you're determining the rate of change of the function at that particular value of \( x \).
In our case, you substitute \( x = 1 \) into the derivative function \( \frac{5}{2}x^{3/2} \) to get \( \frac{5}{2}(1)^{3/2} = \frac{5}{2} \).
Evaluating the derivative provides important information about the behavior of a function, such as its slope, at that specific location.

• Think of it as zooming in on the function graph at \( x = 1 \) to see how steep it is there.
• It also tells us whether the function is increasing or decreasing at \( x = 1 \).

By interpreting the derivative value, we can make conclusions about the function's dynamics at specific points.