Problem 219 - 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: \(\frac{d y}{d x}=\frac{d y}{d u} \frac{d u}{d x}\). $$ y=\sqrt{4 u+3}, u=x^{2}-6 x $$

Step-by-Step Solution

Answer

\( \frac{dy}{dx} = \frac{2(2x - 6)}{\sqrt{4x^2 - 24x + 3}} \)

Step 1: Identify Functions

First, identify the given functions. We have two functions: \( y = \sqrt{4u + 3} \) and \( u = x^2 - 6x \). Here, \( y \) is a function of \( u \), and \( u \) is a function of \( x \).

Step 2: Differentiate First Function with Respect to u

Find the derivative of \( y = \sqrt{4u + 3} \) with respect to \( u \). Use the chain rule: \( \frac{dy}{du} = \frac{1}{2\sqrt{4u + 3}} \times 4 = \frac{2}{\sqrt{4u + 3}} \).

Step 3: Differentiate Second Function with Respect to x

Now, find the derivative of \( u = x^2 - 6x \) with respect to \( x \). Use basic differentiation rules: \( \frac{du}{dx} = 2x - 6 \).

Step 4: Apply Chain Rule Formula

Apply the chain rule formula \( \frac{dy}{dx} = \frac{dy}{du} \times \frac{du}{dx} \). Substitute \( \frac{dy}{du} = \frac{2}{\sqrt{4u + 3}} \) and \( \frac{du}{dx} = 2x - 6 \) into the formula. This gives us \( \frac{dy}{dx} = \frac{2}{\sqrt{4(x^2 - 6x) + 3}} \times (2x - 6) \).

Step 5: Simplify the Expression

Combine and simplify the expression: \( \frac{dy}{dx} = \frac{2(2x - 6)}{\sqrt{4x^2 - 24x + 3}} \).

Key Concepts

Leibniz's notation

In calculus, Leibniz's notation plays a crucial role in expressing derivatives clearly and systematically. When we deal with complex functions, especially those involving multiple variables, this notation becomes even more valuable. The symbol \( \frac{d}{dx} \) represents the derivative of a function with respect to \( x \). It's like saying, "Let's see how this function changes as \( x \) changes."

For functions that depend on other variables, like in the problem \( y = f(u) \) and \( u = g(x) \), Leibniz's notation helps us express these relationships neatly. We have \( \frac{dy}{du} \) for the rate at which \( y \) changes concerning \( u \) and \( \frac{du}{dx} \) for \( u \) with respect to \( x \). With the chain rule, these get combined to find \( \frac{dy}{dx} \), indicating how \( y \) ultimately changes as \( x \) changes.

• The chain rule in Leibniz's notation is expressed as: \( \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} \).
• This clearly tracks the change as it moves through the composite function layers.

Differentiation

Differentiation is the core concept of calculus that deals with finding the rate at which one quantity changes concerning another. It's like determining the speed of a car; knowing how fast it moves is like knowing the derivative.

To differentiate, we use rules and formulas that simplify the process. In the original exercise, basic differentiation rules, such as the power rule and constant rule, were applied. For the function \( u = x^2 - 6x \), using the power rule gives us \( \frac{du}{dx} = 2x - 6 \).

• Power Rule: If \( u = x^n \), then \( \frac{du}{dx} = nx^{n-1} \).
• Constant Factor Rule: If \( u = c \times f(x) \), then \( \frac{du}{dx} = c \times \frac{df}{dx} \) where \( c \) is a constant.

Mastering these rules allows for smooth and accurate calculation of derivatives, which is foundational in solving many calculus problems.

Composite functions

Composite functions occur when one function is nested inside another, like the layers of an onion. For example, if \( y = \sqrt{4u + 3} \) and \( u = x^2 - 6x \), then \( y = f(g(x)) \) is a composite function where \( y \) depends indirectly on \( x \) through \( u \).

Understanding these functions is key to applying the chain rule, as each layer of function impacts the overall derivative process. The formula \( \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} \) reflects this layering, indicating how the changes in \( x \) influence \( y \).

• Identify each function's role: "inner" and "outer" functions are determined by their position in the composition.
• Use the chain rule to handle these layers one step at a time.
• Keep practice to become comfortable at identifying and differentiating composite functions.

Mastering composite functions allows you to tackle more complex calculus problems efficiently and accurately.