Problem 225 - Calculus Volume 1

Question

For the following exercise, 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

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

Step 1: Identify the outer function

The given function is \( y = \csc(\pi x + 1) \). To decompose it, first identify the outer function, which is the function of \( u \). Here, the outer function is \( y = f(u) = \csc(u) \).

Step 2: Identify the inner function

Next, identify the inner function which is expressed in terms of \( x \). The expression inside the cosecant is \( \pi x + 1 \), so the inner function is \( u = g(x) = \pi x + 1 \).

Step 3: Differentiate the outer function with respect to u

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

Step 4: Differentiate the inner function with respect to x

Differentiate the inner function \( u = \pi x + 1 \) with respect to \( x \). The derivative is \( \frac{d}{dx}[\pi x + 1] = \pi \).

Step 5: Apply the chain rule

To find \( \frac{dy}{dx} \), apply the chain rule: \( \frac{dy}{dx} = \frac{dy}{du} \times \frac{du}{dx} \). Substituting the derivatives found, we get \( \frac{dy}{dx} = -\csc(u)\cot(u) \times \pi \).

Step 6: Substitute u back in terms of x

Replace \( u \) with the original inner function \( \pi x + 1 \) to express the derivative in terms of \( x \). Therefore, \( \frac{dy}{dx} = -\pi \csc(\pi x + 1) \cot(\pi x + 1) \).

Key Concepts

Chain Rule

The Chain Rule is a fundamental method in calculus used to find the derivative of a composite function. Imagine a function nested inside another, like a Matryoshka doll. The key is to differentiate both the outer and inner layers separately, then combine these derivatives using multiplication.

Here's how it works:

• Identify both the outer and inner functions. In our example, the outer function was the cosecant, noted as \(y = \csc(u)\), and the inner function was \(u = \pi x + 1\).
• Find the derivative of the outer function with respect to the inner function \(u\). For \(y = \csc(u)\), this derivative is \(-\csc(u)\cot(u)\).
• Next, differentiate the inner function \(u\) with respect to \(x\). In our example, this was just \(\frac{d}{dx}[\pi x + 1] = \pi\).
• Finally, multiply these derivatives together: \( \frac{dy}{dx} = \frac{dy}{du} \times \frac{du}{dx} \).

Applying the chain rule to our exercise simplified the process of finding \(\frac{dy}{dx}\). It allows us to neatly express several layers of functions into a single differentiated form.

Trigonometric Functions

Trigonometric functions, like sine, cosine, and cosecant, play a huge role in calculus. When dealing with a function such as \(y = \csc(\pi x + 1)\), it's crucial to understand their derivatives. Cosecant, which is the reciprocal of sine, has a derivative that uses other trigonometric components like itself and cotangent.

To differentiate \(\csc(u)\), remember:

• Cosecant's derivative is: \( \frac{d}{du}[\csc(u)] = -\csc(u)\cot(u) \).
• This involves both cosecant and cotangent. Such derivatives become particularly meaningful when applying the chain rule.

Cosecant functions often appear in periodic-related calculations or problems dealing with oscillations. In engineering and physics, understanding their differentiation helps in analyzing wave motions and other periodic behaviors. Mastering these can significantly aid in tackling trigonometric-embedded calculus problems.

Differentiation Techniques

Differentiation is all about finding the rate of change, and there are various techniques to address different types of functions.

• Power Rule: Simple polynomials, like \(x^2\), can be tackled using the power rule. Here, the exponent comes down as a coefficient, and you subtract one from the exponent, e.g., \(\frac{d}{dx}[x^n] = nx^{n-1}\).
• Product and Quotient Rules: When a function is a product or a quotient of two simpler functions, these rules let you differentiate efficiently. For example, for a product \(f(x)g(x)\), the derivative is \(f'g + fg'\).
• Chain Rule: As seen above, this is vital for composite functions and requires differentiating both the outer and inner functions separately.

In our exercise, the focus was on the chain rule due to the composition of \(\csc\) and a linear function, but knowing all these techniques provides a solid toolkit for tackling various calculus problems. Differentiation is a powerful tool in mathematical analysis, helping us understand how changes in input affect outputs across numerous domains.