Problem 30 - Calculus Single Variable

Question

Compute \((f \circ g)^{\prime}\) and \((g \circ f)^{\prime}\). $$ f(x)=\sin (x), g(x)=3 x^{5} $$

Step-by-Step Solution

Answer

\((f \circ g)'(x) = 15x^4 \cos(3x^5)\), \((g \circ f)'(x) = 45 \sin^4(x) \cos(x)\).

Step 1: Understand the Composite Function

The exercise involves computing the derivative of composite functions. We have two functions, \(f(x) = \sin(x)\) and \(g(x) = 3x^5\). We are asked to find the derivatives of \((f \circ g)(x) = f(g(x))\) and \((g \circ f)(x) = g(f(x))\).

Step 2: Find \((f \circ g)(x)\)

Substitute \(g(x)\) into \(f(x)\). This gives us \((f \circ g)(x) = \sin(3x^5)\).

Step 3: Differentiate \((f \circ g)(x)\) using the Chain Rule

Using the chain rule, \( (f \circ g)^{\prime}(x) = f^{\prime}(g(x)) \cdot g^{\prime}(x) \). First, find \( f^{\prime}(x) = \cos(x) \) and \( g^{\prime}(x) = 15x^4 \). Therefore, \( (f \circ g)^{\prime}(x) = \cos(3x^5) \cdot 15x^4 \).

Step 4: Substitute and Simplify the Expression

Plug in \(g(x) = 3x^5\) into the derivative formula: \[ (f \circ g)^{\prime}(x) = \cos(3x^5) \cdot 15x^{4} = 15x^4 \cos(3x^5) \].

Step 5: Find \((g \circ f)(x)\)

Substitute \(f(x)\) into \(g(x)\). This gives us \((g \circ f)(x) = 3(\sin(x))^5\).

Step 6: Differentiate \((g \circ f)(x)\) using the Chain Rule

Using the chain rule, \( (g \circ f)^{\prime}(x) = g^{\prime}(f(x)) \cdot f^{\prime}(x) \). Here, \( g^{\prime}(y) = 15y^4 \) and \( f^{\prime}(x) = \cos(x) \). Thus, \( (g \circ f)^{\prime}(x) = 15 (\sin(x))^4 \cos(x) \cdot 1 \cdot 3 \).

Step 7: Substitute and Simplify the Expression

Substituting \(f(x) = \sin(x)\) into the derivative formula, we get \((g \circ f)^{\prime}(x) = 45 \sin^4(x) \cos(x)\).

Key Concepts

Understanding Composite Functions

A composite function involves applying one function to the results of another. In this exercise, we have two functions: \( f(x) = \sin(x) \) and \( g(x) = 3x^5 \). The notation \((f \circ g)(x)\) means we first apply \(g\) to \(x\), and then \(f\) to the result of that application. In other words, we compute \( f(g(x)) \). Likewise, \((g \circ f)(x)\) implies computing \( g(f(x)) \).

When working with composite functions, the order in which you apply the functions matters greatly. For example, \((f \circ g)(x) = \sin(3x^5)\) places \(3x^5\) inside the sine function's input. On the other hand, \((g \circ f)(x) = 3(\sin(x))^5\) involves raising sine to the fifth power inside a multiplication by three.

This concept of composite functions is foundational in calculus as it is often necessary to understand function behavior when two processes are nested into each other.

Finding Derivatives Using the Chain Rule

In calculus, a derivative represents the rate of change of a function with respect to its variable. When dealing with composite functions, we use a technique called the chain rule to find these derivatives. The chain rule states that if you have a composite function \( h(x) = f(g(x)) \), the derivative \( h'(x) \) is calculated as \( f'(g(x)) \cdot g'(x) \).

Let's take \((f \circ g)(x) = \sin(3x^5)\) as an example. To differentiate this, follow these steps:

• Find the derivative of \(f(x) = \sin(x)\), which is \(f'(x) = \cos(x)\).
• Next, find the derivative of \(g(x) = 3x^5\), which is \(g'(x) = 15x^4\).
• Apply the chain rule: \((f \circ g)'(x) = \cos(3x^5) \cdot 15x^4\.\)

This derivative tells you how \((f \circ g)(x)\) changes as \(x\) changes.

Similarly, for \((g \circ f)(x) = 3(\sin(x))^5\), use the chain rule:

• Find \(g'(y) = 15y^4\) where \(y = \sin(x)\).
• Find \(f'(x) = \cos(x)\).
• Combine: \((g \circ f)'(x) = 45 \sin^4(x) \cos(x)\).

This gives you the rate of change for the nested function \(g(f(x))\).

Exploring Trigonometric Functions in Calculus

Trigonometric functions like sine, cosine, and tangent are fundamental in calculus, often describing periodic phenomena. They have specific characteristics that make them unique. For instance, \( \sin(x)\) outputs a value between -1 and 1 for any input \(x\), creating a smooth wave pattern.

When we study trigonometric functions in calculus, calculating their derivatives helps us understand how rapidly these functions increase or decrease at a point. For example, the derivative of \( \sin(x) \) is \( \cos(x) \), giving the slope of the sine function at any point \(x\). This tells us whether \( \sin(x)\) is increasing or decreasing.

Trigonometric functions are also key players in solving real-world problems involving oscillations, such as sound waves or tides. By applying calculus, particularly derivatives, we gain insights into the nature of changes happening within these functions. Understanding these concepts lets us efficiently predict behavior in various scientific and engineering applications. This exercise involves these functions to demonstrate how they can be transformed and analyzed through differentiation.