Problem 1 - Thomas Calculus

Question

In Exercises \(1-8,\) given \(y=f(u)\) and \(u=g(x),\) find \(d y / d x=\) \(f^{\prime}(g(x)) g^{\prime}(x)\). $$ y=6 u-9, \quad u=(1 / 2) x^{4} $$

Step-by-Step Solution

Answer

\( \frac{dy}{dx} = 12x^3 \)

Step 1: Understand the Functions

Given the problem, we have two functions: \( y = f(u) = 6u - 9 \) and \( u = g(x) = \frac{1}{2}x^4 \). Our task is to find the derivative of \( y \) with respect to \( x \), which can be determined using the chain rule.

Step 2: Apply the Chain Rule Formula

The chain rule states that \( \frac{dy}{dx} = f'(g(x)) \cdot g'(x) \). We need to find \( f'(u) \) and \( g'(x) \) to use this formula.

Step 3: Differentiate the Outer Function

The function \( f(u) = 6u - 9 \). The derivative of \( f(u) \) with respect to \( u \) is \( f'(u) = 6 \) because the derivative of a constant is 0.

Step 4: Differentiate the Inner Function

The function \( g(x) = \frac{1}{2}x^4 \). Differentiate this with respect to \( x \) to find \( g'(x) \). Using the power rule, \( g'(x) = 2x^3 \).

Step 5: Substitute into the Chain Rule

Substitute \( f'(u) = 6 \) and \( u = \frac{1}{2}x^4 \) into the chain rule to get \( f'(g(x)) = 6 \cdot 1 = 6 \). Now, multiply this by \( g'(x) = 2x^3 \).

Step 6: Compute the Final Derivative

Combine the results using the chain rule: \( \frac{dy}{dx} = 6 \cdot 2x^3 = 12x^3 \).

Key Concepts

Understanding Derivatives

Derivatives are a fundamental concept in calculus, expressing how a function changes as its input changes. They provide us the rate of change or the slope of the function at any given point. To find the derivative of a function, we need to understand how each part of the function contributes to its overall behavior.
For simpler functions, like polynomials, finding derivatives can be straightforward using standard rules such as the power rule. However, when dealing with compositions of functions, the process becomes slightly more complex and typically involves techniques like the Chain Rule.

Exploring Function Composition

Function composition involves combining two or more functions such that the output of one function becomes the input of the next. In this exercise, we have a clear composition: the function \( y = f(u) = 6u - 9 \) is composed with \( u = g(x) = \frac{1}{2}x^4 \). Here, \( u \) is the intermediary variable linking \( g(x) \) to \( f(u) \).
To differentiate these compositions, we can't simply apply standard rules directly, as the relationship between \( y \) and \( x \) is not explicit. This necessitates the use of the Chain Rule, which helps unravel these layers of functions by relating their derivatives.

Using Differentiation Techniques

Differentiation techniques involve specific methods to find derivatives accurately. One of the most powerful techniques is the Chain Rule, especially useful when dealing with function compositions.
The Chain Rule allows us to differentiate a composite function by taking the derivative of the outer function and multiplying it by the derivative of the inner function. In the solution provided:

• We differentiated the outer function \( f(u) = 6u - 9 \) to get \( f'(u) = 6 \), since the derivative of a constant term is zero, and the derivative of \( 6u \) with respect to \( u \) yields \( 6 \).
• Next, we differentiated the inner function \( g(x) = \frac{1}{2}x^4 \) using the power rule, yielding \( g'(x) = 2x^3 \), since the derivative \( x^n \) is \( nx^{n-1} \).

Combining these, the Chain Rule expression \( \frac{dy}{dx} = f'(g(x)) \cdot g'(x) \), gives us the final result of \( 12x^3 \). This example demonstrates how carefully applying the Chain Rule can unlock complex derivatives of composed functions.