Problem 4 - Calculus - AP Edition

Question

Let \(f(x)=5 \sqrt{x}\) and \(g(x)=4+\cos x\) (a) Find \((f \circ g)(x)\) and \((f \circ g)^{\prime}(x)\) (b) Find \((g \circ f)(x)\) and \((g \circ f)^{\prime}(x)\)

Step-by-Step Solution

Answer

(a) \((f \circ g)(x) = 5 \sqrt{4 + \cos x}\), \((f \circ g)^{\prime}(x) = -\frac{5\sin x}{2\sqrt{4 + \cos x}}\); (b) \((g \circ f)(x) = 4 + \cos(5\sqrt{x})\), \((g \circ f)^{\prime}(x) = -\frac{5\sin(5\sqrt{x})}{2\sqrt{x}}\).

Step 1: Compose Functions

To find \((f \circ g)(x)\), we need to substitute \(g(x)\) into \(f(x)\). Since \(f(x) = 5\sqrt{x}\) and \(g(x) = 4 + \cos x\), \((f \circ g)(x) = f(g(x)) = f(4 + \cos x) = 5 \sqrt{4 + \cos x}\).

Step 2: Differentiate the Composition

To find \((f \circ g)^{\prime}(x)\), we use the chain rule. First, find the derivative of \(f(u) = 5\sqrt{u}\) with respect to \(u\), which is \(\frac{5}{2\sqrt{u}}\). Now, differentiate \(g(x) = 4 + \cos x\), which gives \(-\sin x\). By the chain rule, the derivative is \((f \circ g)^{\prime}(x) = \frac{5}{2\sqrt{4+\cos x}} \times (-\sin x) = -\frac{5\sin x}{2\sqrt{4 + \cos x}}\).

Step 3: Compose the Other Way

To find \((g \circ f)(x)\), we substitute \(f(x)\) into \(g(x)\). So \((g \circ f)(x) = g(f(x)) = g(5\sqrt{x}) = 4 + \cos(5\sqrt{x})\).

Step 4: Differentiate the Other Composition

To find \((g \circ f)^{\prime}(x)\), apply the chain rule again. The derivative of \(g(u) = 4 + \cos u\) with respect to \(u\) is \(-\sin u\). The derivative of \(f(x) = 5\sqrt{x}\) is \(\frac{5}{2\sqrt{x}}\). Thus, \((g \circ f)^{\prime}(x) = -\sin(5\sqrt{x}) \times \frac{5}{2\sqrt{x}} = -\frac{5\sin(5\sqrt{x})}{2\sqrt{x}}\).

Key Concepts

Understanding the Chain Rule

The chain rule is an essential concept in calculus, especially when dealing with composite functions. It allows us to differentiate functions composed of other functions. Think of it like peeling layers of an onion, where each layer is a different function. When applying the chain rule, we differentiate from the outside layer inwards.
This process starts with differentiating the outer function while leaving the inner function unchanged, and then multiplying by the derivative of the inner function.

• First, determine the outer and inner functions. For instance, in a composition like \(f(g(x))\), \(f(u)\) is the outer function and \(g(x)\) is the inner function.
• Differentiate the outer function with respect to its argument. In our problem, for \(f(u) = 5\sqrt{u}\), the derivative is \(\frac{5}{2\sqrt{u}}\).
• Differentiate the inner function and multiply. For \(g(x) = 4 + \cos x\), the derivative is \(-\sin x\).

Connect these calculated derivatives according to the chain rule to find \(\frac{d}{dx}f(g(x)) = f'(g(x)) \cdot g'(x)\), resulting in \(-\frac{5\sin x}{2\sqrt{4 + \cos x}}\) for our exercise.
This is a classic application of the chain rule when dealing with complex differentiations.

Basics of Differentiation

Differentiation is a fundamental concept in calculus used to find the rate at which a quantity changes. It involves finding the derivative of a function, which represents this rate of change. Differentiation provides valuable insight into the behavior of functions, like slopes and tangent lines.
Performing differentiation is straightforward for basic functions, but it becomes interesting when dealing with composite functions, where techniques like the chain rule become crucial.

• For simple power functions like \(f(x) = x^n\), the derivative is \(nx^{n-1}\).
• When dealing with trigonometric functions like \(\cos x\), its derivative is \(-\sin x\).
• Differentiation is linear, meaning the derivative of a sum like \(f(x) = a + b\) is simply the sum of derivatives of \(a\) and \(b\).

In the exercise, applying differentiation allowed us to find derivatives like \(-\sin x\) and \(\frac{5}{2\sqrt{x}}\). These steps are foundational for understanding how changes in one variable affect another in a mathematical model.

Composing Functions

Function composition involves creating a new function by combining two functions. This is done by applying one function to the results of another. In simpler terms, imagine saying what's "inside" another mathematical function. This concept is crucial in calculus, enabling us to build complex functions from simpler ones.
It's important to understand how to compose functions correctly:

• The composition \((f \circ g)(x)\) means you first apply \(g(x)\) and then \(f\). In our exercise, this resulted in \(f(g(x)) = 5 \sqrt{4 + \cos x}\).
• Reversing the composition, as in \((g \circ f)(x)\), means applying \(f(x)\) first, resulting in \(g(f(x)) = 4 + \cos(5 \sqrt{x})\).

Composition is like nesting layers, and each layer affects the subsequent ones. Recognizing this relationship helps in understanding not only calculus problems but real-world phenomena represented by these functions. Understanding function composition and its applications lays the groundwork for more advanced calculus concepts.