Problem 2 - Thomas Calculus

Question

In Exercises \(1-8,\) given \(y=f(u)\) and \(u=g(x),\) find \(d y / d x=\) \(d y / d x=f^{\prime}(g(x)) g^{\prime}(x)\) $$ y=2 u^{3}, \quad u=8 x-1 $$

Step-by-Step Solution

Answer

The derivative is \( \frac{dy}{dx} = 48(8x - 1)^2 \).

Step 1: Differentiate Inner Function

First, we identify the inner function as \(u = 8x - 1\). The derivative of this function with respect to \(x\) is calculated as follows:\[ g'(x) = \frac{d}{dx}(8x - 1) = 8. \]

Step 2: Differentiate Outer Function

Next, differentiate the outer function \(y = 2u^3\) with respect to \(u\). The derivative will be:\[ f'(u) = \frac{d}{du}(2u^3) = 6u^2. \]

Step 3: Apply Chain Rule

To find \(\frac{dy}{dx}\), apply the chain rule which states: \( \frac{dy}{dx} = f'(g(x)) \cdot g'(x) \). We substitute \(g(x)\) back in and perform the calculation:\[ \frac{dy}{dx} = f'(u) \cdot g'(x) = 6u^2 \cdot 8. \]

Step 4: Substitute for u

Finally, substitute back \(u = 8x - 1\) into the expression we obtained in Step 3:\[ \frac{dy}{dx} = 6(8x - 1)^2 \times 8. \]This gives us the derivative, \( \frac{dy}{dx} = 48(8x - 1)^2 \).

Key Concepts

Derivative of Composite Functions

When dealing with **composite functions**, finding their derivative can seem intimidating at first. However, with the right approach and a solid understanding of the chain rule, it becomes quite manageable. A composite function is formed when you have a function within a function, like our example where we have \( y = f(u) \) and \( u = g(x) \). To find the derivative of such functions, we use the chain rule:

• First, identify the inner function \( u = g(x) \)
• Then, the outer function \( y = f(u) \)

The chain rule provides a powerful method to differentiate these composite structures by expressing \( \frac{dy}{dx} \) in terms of derivatives of the inner and outer functions. It tells us that the derivative of a composite function \( y = f(g(x)) \) is the product of the derivative of the outer function evaluated at the inner function, multiplied by the derivative of the inner function. This is compactly written as:\[\frac{dy}{dx} = f'(g(x)) \cdot g'(x)\]By understanding this, you simplify the process of finding derivatives of complex nested functions.

Inner Function Differentiation

The **inner function** is the one inside when you look at composite functions like \( y = f(u) = 2u^3 \) and \( u = g(x) = 8x - 1 \). Here, the inner function is \( u = 8x - 1 \), and differentiating it is straightforward. When you differentiate the inner function with respect to \( x \), you're essentially finding out how sensitive \( u \) is to changes in \( x \).
To differentiate \( u = 8x - 1 \), we find:

• The derivative \( g'(x) = \frac{d}{dx}(8x - 1) \)
• This gives us a simple result: \( g'(x) = 8 \)

This derivative tells us that for each unit increase in \( x \), \( u \) increases by 8 units. This result will be an important factor as we apply the chain rule in the next steps.

Outer Function Differentiation

The **outer function** in our problem is \( y = 2u^3 \), which is dependent on the inner function \( u \). Differentiating the outer function means calculating the derivative with respect to \( u \), taking its form into account.
Let's break it down:

• The outer function is \( y = 2u^3 \)
• To differentiate with respect to \( u \), apply the power rule: \( \frac{d}{du}(2u^3) = 6u^2 \)

This result \( f'(u) = 6u^2 \) showcases how the outer function's rate of change depends on \( u \). When you take the derivative of \( y \) with respect to \( u \), you're essentially looking at how \( y \) changes as \( u \) changes. It's crucial to substitute \( u \) back into the original equation for final calculations, as seen when applying the chain rule.