Problem 226 - Calculus Volume 1

Question

For each of the following exercises, a. decompose each function in the form \(y=f(u)\) and \(u=g(x),\) and b. find \(\frac{d y}{d x}\) as a function of \(x .\) $$y=\cot ^{2} x$$

Step-by-Step Solution

Answer

\( \frac{dy}{dx} = -2 \cot{x} \csc^2{x} \)

Step 1: Decompose y into f(u) and u = g(x)

To decompose the function, identify the inner function and the outer function. Here, rewrite the given function in a different form:1. **Outer Function**: Consider the square as the outer function, so let \( y = u^2 \).2. **Inner Function**: Let \( u = \cot{x} \).Thus, we have \( y = f(u) = u^2 \) and \( u = g(x) = \cot{x} \).

Step 2: Differentiate the outer function with respect to u

Find the derivative of the outer function \( f(u) = u^2 \) with respect to \( u \):\[ \frac{dy}{du} = \frac{d}{du}(u^2) = 2u \]

Step 3: Differentiate the inner function with respect to x

Find the derivative of the inner function \( u = \cot{x} \) with respect to \( x \):\[ \frac{du}{dx} = \frac{d}{dx}(\cot{x}) = -\csc^2{x} \]

Step 4: Apply the Chain Rule

Use the chain rule to find \( \frac{dy}{dx} \). According to the chain rule:\[ \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} \]Substitute the derivatives from Steps 2 and 3:\[ \frac{dy}{dx} = (2u) \left(-\csc^2{x}\right) \]

Step 5: Substitute back the inner function u

Substitute back \( u = \cot{x} \) into the expression for \( \frac{dy}{dx} \):\[ \frac{dy}{dx} = 2 \cot{x} \cdot (-\csc^2{x}) \]Simplify the expression:\[ \frac{dy}{dx} = -2 \cot{x} \csc^2{x} \]

Step 6: Conclusion

The derivative of the function \( y = \cot^2{x} \) with respect to \( x \) is \( \frac{dy}{dx} = -2 \cot{x} \csc^2{x} \).

Key Concepts

Chain Rule

The chain rule is an essential technique in calculus for differentiating composite functions. When a function is composed of an "outer" function affecting an "inner" one, the derivative of the whole is found by differentiating the outer function and then multiplying it by the derivative of the inner function, like nesting dolls. Imagine two linked machines, where one machine's output is fed into another. To model this in mathematics, if you have functions defined as \( y = f(u) \) and \( u = g(x) \), the chain rule helps you find the rate of change of \( y \), with respect to \( x \), using the formula:

• \( \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} \)

This formula encapsulates the idea that because the effect on \( y \) passes through \( u \), both rates of change are needed. In our example, the outer function is the square operation, \( y = u^2 \), and the inner function is \( u = \cot{x} \). So, differentiating \( y \) with respect to \( u \) first, and then \( u \) with respect to \( x \), allows us to get \( \frac{dy}{dx} \).

Derivative of Trigonometric Functions

Trigonometric functions are a staple in calculus, frequently appearing in calculations involving periodic phenomena. Among such functions, the cotangent, represented as \( \cot{x} \), is just one example. Finding derivatives for these functions requires knowing the basic derivative rules:

• The derivative of \( \sin{x} \) is \( \cos{x} \)
• The derivative of \( \cos{x} \) is \( -\sin{x} \)
• The derivative of \( \cot{x} \) is \( -\csc^2{x} \)

These derivatives stem from the fundamental properties of trigonometric functions and their geometric implications on a circle. In this problem, understanding that the derivative of \( \cot{x} \) is \( -\csc^2{x} \) was vital for applying the chain rule correctly. Recognizing these derivatives will help you compute the rate of change in any number of scenarios involving cyclic behaviors.

Function Decomposition

Function decomposition is the process of breaking a composite function into its basic parts, usually an inner and outer function. It assists in understanding the structure of complex expressions and is especially useful when applying the chain rule. In simple terms:

• Identify the so-called outer function which works on an intermediate result from the inner function.
• Determine the inner function that transforms the initial input.

For example, in the exercise \( y = \cot^2{x} \), the decomposition process reveals \( y = f(u) = u^2 \) as the outer function, and \( u = g(x) = \cot{x} \) as the inner one. By dissecting functions like this, it becomes easier to handle differentiation because each piece can be considered separately before recombining. This strategy underlies a lot of differentiation work done in calculus.