Problem 215 - Calculus Volume 1

Question

Given \(y=f(u)\) and \(u=g(x),\) find \(\frac{d y}{d x}\) by using Leibniz's notation for the chain rule: \(\frac{d y}{d x}=\frac{d y}{d u} \frac{d u}{d x}\). $$ y=6 u^{3}, u=7 x-4 $$

Step-by-Step Solution

Answer

\( \frac{d y}{d x} = 126(7x - 4)^2 \).

Step 1: Recognize the Functions

We have the functions given as \( y = f(u) = 6u^3 \) and \( u = g(x) = 7x - 4 \). We will differentiate each with respect to their respective variables.

Step 2: Differentiate y with respect to u

First, differentiate \( y = 6u^3 \) with respect to \( u \) using the power rule. The power rule states that the derivative of \( au^n \) is \( anu^{n-1} \). Thus, \( \frac{dy}{du} = 6 \times 3u^{2} = 18u^{2} \).

Step 3: Differentiate u with respect to x

Next, differentiate \( u = 7x - 4 \) with respect to \( x \). The derivative of \( 7x \) is \( 7 \) and the derivative of any constant is \( 0 \). Thus, \( \frac{du}{dx} = 7 \).

Step 4: Apply the Chain Rule

Apply Leibniz's notation for the chain rule: \( \frac{dy}{dx} = \frac{dy}{du} \times \frac{du}{dx} \). Substitute the derivatives found: \( \frac{dy}{dx} = 18u^{2} \times 7 = 126u^{2} \).

Step 5: Substitute u back in terms of x

Substitute \( u = 7x - 4 \) into \( 126u^{2} \) to express the derivative in terms of \( x \). Thus, \( \frac{dy}{dx} = 126(7x - 4)^{2} \).

Key Concepts

Leibniz notation

Leibniz notation is a way of expressing derivatives that makes the process of differentiation clear and organized. Named after the mathematician Gottfried Wilhelm Leibniz, this notation uses the "/" symbol to signify the rate of change between two variables. For example, \( \frac{dy}{dx} \) represents the derivative of \( y \) with respect to \( x \). This is particularly useful in applications of the chain rule, which deals with composite functions.
In our exercise, we used Leibniz notation to express the process of finding the derivative of \( y \) with respect to \( x \), when \( y \) depends on \( u \, \text{and} \ u \) depends on \( x \). The chain rule, in Leibniz notation, is written as \( \frac{dy}{dx} = \frac{dy}{du} \times \frac{du}{dx} \). This formula serves as a guide for linking the rate of change of \( y \) with respect to \( x \) through the intermediary variable \( u \).
By applying this notation, you clearly see each step in the process, making it easier to track changes and solve more complex differentiation problems straightforwardly.

Differentiation

Differentiation is the process of finding the derivative of a function. The derivative represents the rate at which a function value changes as its input changes. It is a fundamental tool in calculus and helps define concepts such as slope, velocity, and acceleration.
In the problem provided, we perform differentiation in two parts. First, we differentiate \( y = 6u^3 \) with respect to \( u \), giving us \( \frac{dy}{du} = 18u^2 \). This tells us how \( y \) changes as \( u \) changes. Second, we differentiate \( u = 7x - 4 \) with respect to \( x \), resulting in \( \frac{du}{dx} = 7 \). This indicates how \( u \) changes as \( x \) changes.
By combining these derivatives through the chain rule, we discover how \( y \) changes with respect to \( x \), encapsulating the interplay of direct and indirect changes between these variables. Mastery of differentiation skills is essential for handling more advanced calculus problems.

Power Rule

The power rule is a basic and crucial technique used in differentiation to find derivatives of functions in the form of a power, like \( ax^n \). According to this rule, the derivative of \( ax^n \) with respect to \( x \) is given by multiplying the exponent \( n \) by the coefficient \( a \,\text{and} \ \) then reducing the exponent by one to get \( anx^{n-1} \).
During our exercise, we applied the power rule to \( y = 6u^3 \). We took the exponent \( 3 \,\text{and} \ \) multiplied it by \( 6 \,\text{resulting in} \ \) \( 18u^2 \). This simple step is the bread-and-butter of many calculus problems, allowing us to quickly find slopes of tangents and rates of change for polynomial expressions.
Learning the power rule is integral (pun intended!) to progressing through calculus because it extends beyond mere polynomial expressions to more complex functions, building a strong foundational understanding in calculus.