Problem 28 - Calculus Single Variable

Question

Compute \((f \circ g)^{\prime}\) and \((g \circ f)^{\prime}\). $$ f(x)=3 x^{5}-2 x^{2}, g(x)=\sin (x) $$

Step-by-Step Solution

Answer

\((f \circ g)'(x) = (15 \sin^4(x) - 4 \sin(x)) \cos(x)\) and \((g \circ f)'(x) = \cos(3x^5 - 2x^2) \cdot (15x^4 - 4x)\).

Step 1: Understand the problem

We are asked to compute the derivatives of two compositions of functions: \((f \circ g)(x)\) and \((g \circ f)(x)\). These relate to finding \((f \circ g)'(x)\) and \((g \circ f)'(x)\) using the chain rule.

Step 2: Apply the chain rule to (f ∘ g)'(x)

Using the chain rule, the derivative \((f \circ g)'(x)\) is computed as \( f'(g(x)) \cdot g'(x) \). We first find the derivatives \( f'(x) \) and \( g'(x) \).

Step 3: Derive f'(x)

For \( f(x) = 3x^5 - 2x^2 \), use the power rule. The derivative is \( f'(x) = 15x^4 - 4x \).

Step 4: Derive g'(x)

For \( g(x) = \sin(x) \), the derivative is \( g'(x) = \cos(x) \).

Step 5: Compute (f ∘ g)'(x)

Substitute the derivatives into the chain rule formula: \( (f \circ g)'(x) = f'(g(x)) \cdot g'(x) = (15(\sin(x))^4 - 4\sin(x)) \cdot \cos(x) \).

Step 6: Apply the chain rule to (g ∘ f)'(x)

Similarly, compute \((g \circ f)'(x) = g'(f(x)) \cdot f'(x)\). Use the derivatives from earlier steps.

Step 7: Compute (g ∘ f)'(x)

Substitute into the chain rule formula: \( (g \circ f)'(x) = \cos(f(x)) \cdot (15x^4 - 4x) \).

Key Concepts

Function Composition

Function composition involves putting one function inside another. With two functions, say \( f(x) \) and \( g(x) \), the composition can be \((f \circ g)(x)\) or \((g \circ f)(x)\).
In simple terms:

• For \((f \circ g)(x)\), input \(x\) into \(g(x)\) first and then into \(f\).
• For \((g \circ f)(x)\), input \(x\) into \(f(x)\) first and then into \(g\).

In our case, understand that you are effectively nesting functions like a stack of boxes, with one function acting as the outer box and the other as the inner box.
This is a foundational concept in calculus, especially when you start applying it to derivatives using the chain rule. As functions become more complex, function composition helps you build and break down expressions into manageable parts.

Derivative

The derivative of a function represents its rate of change. If you picture a curve, the derivative at any point on the curve gives the slope of the tangent line at that point.
For example, given a function \( y = f(x) \), the derivative is often denoted as \( f'(x) \) or \( \frac{dy}{dx} \).
Let's explore some basic rules:

• If \( f(x) = 3x^5 - 2x^2 \), you find \( f'(x) \) to understand how quickly or slowly \( f(x) \) changes at any given \( x \).
• For \( g(x) = \sin(x) \), \( g'(x) = \cos(x) \) informs us how the sine function changes.

Recognizing and calculating derivatives is crucial, especially for function compositions, because it allows us to analyze their behavior step by step and apply rules such as the chain rule.

Power Rule

The power rule is a straightforward method to find the derivative of functions of the form \( x^n \). The derivative of \( x^n \) is given by \( nx^{n-1} \).
For instance, if you have \( f(x) = 3x^5 - 2x^2 \):

• Apply the power rule on \( 3x^5 \): the derivative is \( 15x^4 \).
• Apply the power rule on \( -2x^2 \): the derivative is \( -4x \).

Combine these to get \( f'(x) = 15x^4 - 4x \).
Notice how the power rule quickly helps in simplifying the process of finding derivatives of polynomial functions, making it efficient and easy to apply especially when dealing with function compositions.

Sine Function

The sine function, represented as \( \sin(x) \), is one of the six fundamental trigonometric functions. It describes the y-coordinate of a point on the unit circle, as this angle increases.
A remarkable feature of the sine function is its periodic nature: every \( 2\pi \), the values repeat. For calculus, understanding its derivative is crucial:

• The derivative of \( \sin(x) \) is \( \cos(x) \).

This tells us how the sine function changes as \( x \) changes.
In the context of our exercise, knowing the derivative of the sine function helps to apply the chain rule to compositions involving \( \sin(x) \), such as in \((f \circ g)'(x)\) or \((g \circ f)'(x)\). This interplay between trigonometric and algebraic functions is an important skill in calculus, laying groundwork for more complex analyses of functions.