Problem 82 - Calculus Early Transcendentals

Question

Let \(y=f_{1}(u), u=f_{2}(v), v=f_{3}(w),\) and \(w=f_{4}(x) .\) Express \(d y / d x\) in terms of \(d y / d u, d w / d x, d u / d v,\) and \(d v / d w .\)

Step-by-Step Solution

Answer

\( \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dv} \cdot \frac{dv}{dw} \cdot \frac{dw}{dx} \).

Step 1: Identify the Chain Rule Concept

The function consists of multiple intermediary functions, meaning we can apply the chain rule of differentiation multiple times to express the derivative of the outer function with respect to the innermost variable.

Step 2: Differentiate Inner to Outer

Given the composition of the functions, start with the innermost function and apply the chain rule sequentially to find the expression for \ \( \frac{dy}{dx} \ \).According to the chain rule:\[ \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dv} \cdot \frac{dv}{dw} \cdot \frac{dw}{dx}\]

Step 3: Substitute Given Derivatives

The differentiation results are expressed as:- \( \frac{dy}{du} \) is the derivative of function \( y = f_1(u) \) in terms of \( u \).- \( \frac{dw}{dx} \) is the derivative of function \( w = f_4(x) \) with respect to \( x \).- \( \frac{du}{dv} \) is the derivative of function \( u = f_2(v) \) with respect to \( v \).- \( \frac{dv}{dw} \) is the derivative of function \( v = f_3(w) \) with respect to \( w \).

Step 4: Final Expression

Combine all the individual derivatives according to the chain rule formula obtained in Step 2 to give the expression for \ \( \frac{dy}{dx} \ \):\[ \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dv} \cdot \frac{dv}{dw} \cdot \frac{dw}{dx}\]

Key Concepts

Differentiation

Differentiation is a mathematical process used to find how a function changes as its input changes. It's like figuring out how fast something is moving or how steep a hill is at any point. Differentiation helps us understand the rate of change of a function concerning a variable.

One of the most important tools for differentiation is understanding how different rules apply, such as the power rule, product rule, and chain rule. In our exercise, we use the **Chain Rule** because we are dealing with a function composed of other functions. It's a little like unraveling a series of connected loops, one inside the other.

• The Chain Rule helps in breaking down functions so that we can differentiate them bit by bit.
• This is useful in real-world applications where variables are dependent on each other.

Without differentiation, we would not be able to calculate various real-life movements and changes, from tracking the speed of a car to rates of chemical reactions.

Composite Functions

Composite functions are like functions within functions. Imagine a set of Russian dolls: each doll fits inside the next, and that's how composite functions work. Each function acts as input for the next one until you reach the final output.

In our exercise, we have a chain of functions: \(y=f_1(u)\), \(u=f_2(v)\), \(v=f_3(w)\), and \(w=f_4(x)\). Each output becomes an input for the next. Understanding composite functions is key to applying the chain rule efficiently.

• You need to understand how each function in the chain interacts with others.
• Change in the innermost variable affects all nested functions.
• Composite functions allow us to handle complex relationships in mathematics.

Through this step-by-step unraveling, what seems complicated can become as simple as dealing with each function in sequence.

Derivatives

Derivatives tell us about the rate at which one quantity changes with respect to another. Think of it like understanding how fast something is moving at any given point. In mathematical terms, it often represents the slope or steepness of a curve.

In our exercise, we find the derivative \(\frac{dy}{dx}\) using multiple steps, tackling each function one at a time. The derivative of a composite function involves each small part:

• First, find \(\frac{dy}{du}\), representing how \(y\) changes with \(u\).
• Then, \(\frac{du}{dv}\) shows the change of \(u\) with \(v\).
• Next, \(\frac{dv}{dw}\) deals with the change of \(v\) with \(w\).
• Finally, \(\frac{dw}{dx}\) illustrates the change of \(w\) with \(x\).

By combining all these, we understand how the tiniest change in \(x\) affects the final change in \(y\). This calculation is crucial for understanding incremental changes and making informed predictions about a wide array of natural phenomena.