Problem 217 - 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: \(\frac{d y}{d x}=\frac{d y}{d u} \frac{d u}{d x}\). $$ y=\cos u, u=\frac{-x}{8} $$

Step-by-Step Solution

Answer

\( \frac{dy}{dx} = \frac{1}{8} \sin\left(\frac{-x}{8}\right) \)

Step 1: Define Derivatives

To find \( \frac{dy}{dx} \), we must first determine two intermediary derivatives using the functions given. Start by identifying \( \frac{dy}{du} \) from the function \( y = \cos u \), which gives \( \frac{dy}{du} = -\sin u \).

Step 2: Differentiate Second Function

Next, differentiate \( u = \frac{-x}{8} \) with respect to \( x \), which gives \( \frac{du}{dx} = -\frac{1}{8} \).

Step 3: Apply Chain Rule

Now that we have \( \frac{dy}{du} = -\sin u \) and \( \frac{du}{dx} = -\frac{1}{8} \), apply the chain rule formula: \( \frac{dy}{dx} = \frac{dy}{du} \times \frac{du}{dx} \). Substitute the derivatives: \( \frac{dy}{dx} = -\sin u \times -\frac{1}{8} \).

Step 4: Simplify the Expression

Multiply the expressions obtained: \( \frac{dy}{dx} = \left(-\sin u\right) \times \left(-\frac{1}{8}\right) = \frac{1}{8} \sin u \).

Step 5: Substitute in Terms of x

Replace \( u \) with its expression in terms of \( x \) from \( u = \frac{-x}{8} \). The final derivative \( \frac{dy}{dx} = \frac{1}{8} \sin\left(\frac{-x}{8}\right) \).

Key Concepts

Leibniz's notation in Chain Rule

Leibniz's notation is a clever and versatile way to express derivatives. It uses the symbols \( \frac{d}{d} \) to denote differentiation. In the context of a chain rule, which is used for functions composed of other functions, this notation offers clear guidance and steps for differentiation.
If we have two functions, say \( y = f(u) \) and \( u = g(x) \), we wish to find how \( y \) changes with respect to \( x \). This means finding \( \frac{dy}{dx} \).

• Here, \( \frac{dy}{dx} \) symbolizes the rate of change of \( y \) with \( x \).
• To break it down, \( \frac{dy}{du} \) is the change of \( y \) with \( u \), and \( \frac{du}{dx} \) is the change of \( u \) with \( x \).
• Using Leibniz's notation, chain rule becomes: \( \frac{dy}{dx} = \frac{dy}{du} \times \frac{du}{dx} \).

This makes it easy to follow and solve complex differentiations step-by-step. With our original problem, substituting values into this form allows us to systematically find \( \frac{dy}{dx} \).

Understanding Derivatives

Derivatives measure how a function changes as its input changes. They are fundamental in calculus, helping describe how quantities vary and predicting future behavior. In simpler terms, if you know the derivative, you can understand how two variables interact.
The basics of derivatives include:

• The derivative of a constant is zero because constant values do not change.
• For a linear function \( f(x) = ax \), the derivative \( f'(x) \) is simply \( a \).
• When dealing with the product of functions, we use the product rule.

In the context of our problem, two derivatives are crucial:

• Finding \( \frac{dy}{du} \) from \( y = \cos u \), resulting in \( -\sin u \).
• Finding \( \frac{du}{dx} \) from \( u = \frac{-x}{8} \), resulting in \( -\frac{1}{8} \).

Derivatives not only help in mathematical computations but also in understanding physical and real-world phenomena.

Trigonometric Functions in Calculus

Trigonometric functions like \( \sin \), \( \cos \), and \( \tan \) play a vital role when dealing with periodic phenomena or angles. In calculus, they help describe oscillations, waves, and cyclic behavior.
When differentiating these functions, certain rules apply:

• The derivative of \( \sin x \) is \( \cos x \).
• The derivative of \( \cos x \) is \( -\sin x \).
• Formulas like \( \tan x = \frac{\sin x}{\cos x} \) can sometimes be used.

In our specific example, we differentiate \( y = \cos u \) to obtain \( \frac{dy}{du} = -\sin u \).
These principles can be applied throughout different areas of calculus, simplifying complex functions and making them manageable. Understanding these rules helps unlock the potential of calculus to solve real-world problems involving trigonometric functions.