Problem 225 - 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 \(x .\) $$y=\csc (\pi x+1)$$

Step-by-Step Solution

Answer

\( \frac{dy}{dx} = -\pi \csc(\pi x + 1)\cot(\pi x + 1) \)

Step 1: Decompose the Function

To decompose the given function, identify the outer function and inner function. In the given equation \( y = \csc(\pi x + 1) \), treat \( u = \pi x + 1 \) as the inner function and \( y = \csc(u) \) as the outer function.

Step 2: Differentiate the Inner Function

Differentiate the inner function \( u = \pi x + 1 \) with respect to \( x \). This gives \( \frac{du}{dx} = \pi \).

Step 3: Differentiate the Outer Function

Differentiate the outer function \( y = \csc(u) \) with respect to \( u \). The derivative is \( \frac{dy}{du} = -\csc(u)\cot(u) \).

Step 4: Apply the Chain Rule

To find \( \frac{dy}{dx} \), apply the chain rule, which states that \( \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} \). Substitute the derivatives found in Steps 2 and 3 to get: \( \frac{dy}{dx} = -\csc(\pi x + 1)\cot(\pi x + 1) \cdot \pi \).

Step 5: Simplify the Expression

Combine the terms to express the derivative fully. The solution is \( \frac{dy}{dx} = -\pi \csc(\pi x + 1)\cot(\pi x + 1) \). This represents \( \frac{dy}{dx} \) as a function of \( x \).

Key Concepts

Chain Rule

The chain rule is an essential tool in calculus for differentiating composite functions, which are functions composed of other functions. For example, when you have a function like \( y = \csc(\pi x + 1) \), it means that the output of one function is the input of another.

To apply the chain rule, you first identify the "inner" and "outer" functions. In this instance, the inner function is \( u = \pi x + 1 \) and the outer function is \( y = \csc(u) \).

Once identified, differentiate the inner function with respect to \( x \), giving us \( \frac{du}{dx} = \pi \). Next, differentiate the outer function with respect to \( u \), yielding \( \frac{dy}{du} = -\csc(u)\cot(u) \).

The beauty of the chain rule is in its simplicity: \( \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} \). By substituting the derivatives we found, we calculate \( \frac{dy}{dx} = -\pi \csc(\pi x + 1)\cot(\pi x + 1) \). This formula helps find the rate of change of \( y \) with respect to \( x \) in complex relationships.

Trigonometric Functions

Trigonometric functions are fundamental in mathematics for relating the angles and sides of triangles and oscillatory phenomena like waves. In this context, the function \( \csc(u) \) represents the cosecant, which is reciprocal to the sine function, \( \csc(u) = \frac{1}{\sin(u)} \).

When differentiating trigonometric functions like the cosecant, the derivative involves other trig functions. Specifically, the derivative of \( \csc(u) \) is \( -\csc(u) \cot(u) \). Here, \( \cot(u) \) stands for the cotangent, which is the reciprocal of the tangent function: \( \cot(u) = \frac{\cos(u)}{\sin(u)} \).

These derivative properties are key in calculus and are particularly useful when applying the chain rule. By understanding how these functions interact, you can better predict and compute changes in trigonometric relationships.

Calculus

Calculus is a branch of mathematics that studies continuous change, primarily through derivatives and integrals. Derivatives focus on finding how a function changes at any point, which is perfect for solving problems involving rates of change and curves.

In this exercise, we're dealing with finding the derivative \( \frac{dy}{dx} \) which represents the rate of change of \( y \) with respect to \( x \). The chain rule plays an integral role here, as it allows the handling of composite functions by breaking them down into simpler parts.

By applying calculus principles, you gain a powerful toolset to not only solve algebraic problems but understand the dynamics of a wide range of systems, from motion in physics to the growth rates in biology. Therefore, understanding the foundations of calculus will give you insight into many natural phenomena.