Problem 216 - 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: \(\frac{d y}{d x}=\frac{d y}{d u} \frac{d u}{d x}\). $$ y=\sin u, u=5 x-1 $$

Step-by-Step Solution

Answer

\( \frac{d y}{d x} = 5\cos(5x - 1) \).

Step 1: Recognize the functions

First, identify the given functions and how they relate to each other. We have the function \( y = f(u) = \sin(u) \), and the inner function \( u = g(x) = 5x - 1 \).

Step 2: Differentiate Outer Function with respect to u

Find the derivative of \( y \) with respect to \( u \), which is \( \frac{d y}{d u} \). Given \( y = \sin(u) \), we differentiate using the chain rule for trigonometric functions: \( \frac{d y}{d u} = \cos(u) \).

Step 3: Differentiate Inner Function with respect to x

Find \( \frac{d u}{d x} \) by differentiating \( u = 5x - 1 \) with respect to \( x \). The derivative is \( \frac{d u}{d x} = 5 \).

Step 4: Apply the Chain Rule

Use the chain rule \( \frac{d y}{d x} = \frac{d y}{d u} \cdot \frac{d u}{d x} \) to find \( \frac{d y}{d x} \). Substitute the previously calculated derivatives: \( \frac{d y}{d x} = \cos(u) \cdot 5 \).

Step 5: Substitute Back the Expression for u

Replace \( u \) with the expression in terms of \( x \) from the inner function. Thus, \( \frac{d y}{d x} = 5\cos(5x - 1) \).

Key Concepts

Understanding Derivatives

Derivatives are a fundamental concept in calculus. They measure how a function changes as its input changes. Essentially, a derivative represents the slope of a function at any given point. This slope tells us the rate of change of the function. For a function \( y = f(x) \), the derivative is often denoted as \( \frac{dy}{dx} \).
In the context of this exercise, we're dealing with the composition of two functions: \( y = \sin u \) and \( u = 5x - 1 \). The goal is to find the derivative of \( y \) with respect to \( x \), denoted as \( \frac{dy}{dx} \). Here, the chain rule is essential, as it allows us to differentiate composite functions by relating them through their individual derivatives.

Exploring Trigonometric Functions

Trigonometric functions, like sine and cosine, arise frequently in calculus. They are especially important when dealing with angles and periodic phenomena such as waves. The trigonometric function we're focusing on in this exercise is \( \sin(u) \).
When deriving trigonometric functions, we use specific rules. For instance, the derivative of \( \sin(u) \) with respect to \( u \) is \( \cos(u) \). This pattern proves very useful, particularly when combined with the chain rule, allowing us to differentiate more complex expressions involving trigonometric functions. Here, recognizing these derivatives helps simplify the process and makes solving the exercise more straightforward.

Leibniz's Notation and The Chain Rule

Leibniz's notation helps clearly express the derivatives we calculate. It's especially useful in the context of the chain rule. The chain rule itself is a method for finding the derivative of composite functions. It states: \( \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} \).
This method breaks down the problem into simpler parts, allowing us to first find \( \frac{dy}{du} \) and \( \frac{du}{dx} \), and then multiply these derivatives together. In this exercise, applying the chain rule involves these precise steps:

• Differentiating \( y = \sin(u) \), giving \( \frac{dy}{du} = \cos(u) \).
• Differentiating \( u = 5x - 1 \), where \( \frac{du}{dx} = 5 \).
• Finally, combining these results as \( \frac{dy}{dx} = \cos(u) \cdot 5 \).

Substituting back \( u = 5x - 1 \), we find \( \frac{dy}{dx} = 5\cos(5x - 1) \). This seamless application of Leibniz's notation provides a clear and structured way to tackle derivatives.