Problem 232 - Calculus Volume 1

Question

For the following exercises, find \(\frac{d y}{d x}\) for each function. $$y=\frac{1}{\sin ^{2}(x)}$$

Step-by-Step Solution

Answer

\(-2\frac{\cos(x)}{\sin^3(x)}\)

Step 1: Rewrite the function

First, rewrite the function in a form that is easier to differentiate. The function given is \( y = \frac{1}{\sin^2(x)} \). This can be expressed as \( y = \sin^{-2}(x) \) using exponent rules.

Step 2: Apply the Chain Rule

The chain rule is applied to differentiate \( y = \sin^{-2}(x) \). This means using \( \frac{d}{dx}[u^n] = n \cdot u^{n-1} \cdot \frac{du}{dx} \) where \( u = \sin(x) \) and \( n = -2 \).

Step 3: Differentiate using the Chain Rule

Differentiate \( y = \sin^{-2}(x) \) using the chain rule: \[ \frac{dy}{dx} = -2 \cdot \sin(x)^{-3} \cdot \cos(x) \]This comes from differentiating \( \sin^{-2}(x) \) with respect to \( x \), where \( \frac{d}{dx}[\sin(x)] = \cos(x) \).

Step 4: Simplify the derivative

Simplify the expression for the derivative: \[ \frac{dy}{dx} = -2 \frac{\cos(x)}{\sin^3(x)} \]Thus, \[ \frac{dy}{dx} = -2 \cdot \frac{\cos(x)}{\sin^3(x)} \] or written as \( -2 \cdot \cot(x) \cdot \csc^2(x) \).

Key Concepts

Chain Rule

The chain rule is a fundamental concept in calculus, especially when dealing with composite functions. When you have a function composed of other functions, like in our exercise with the sine function raised to a power, the chain rule helps us differentiate them efficiently.
This rule states that if you have a composite function of the form \( f(g(x)) \), the derivative is calculated as the derivative of \( f \) with respect to \( g \) multiplied by the derivative of \( g \) with respect to \( x \). In symbol form, it’s:

• \( \frac{d}{dx} [f(g(x))] = f'(g(x)) \cdot g'(x) \)

So, if we take our exercise example, which involves \( y = \sin^{-2}(x) \), we can break this down by:

• First differentiating \( u^n \) where \( u = \sin(x) \) and \( n = -2 \).
• Using the formula, we have \( \frac{d}{dx} [u^n] = n \cdot u^{n-1} \cdot \frac{du}{dx} \).

This allows us to effectively find the derivative for complex functions involving multiple operations by decomposing them into more manageable steps.

Sine Function Differentiation

Differentiating trigonometric functions, like sine, forms the basis for handling more complex problems in calculus. For the sine function, the process is straightforward but crucial.
Let's consider the basic derivative of the sine function:\( \sin(x) \). The derivative of \( \sin(x) \) with respect to \( x \) is \( \cos(x) \), meaning that as we differentiate, the sine transforms into a cosine.
This particular property is often used repeatedly, especially in applications like our example where \( y = \sin^{-2}(x) \). During differentiation in the chain rule application:

• The sine part, \( \sin(x) \), becomes \( \cos(x) \) upon differentiation.

As a student, once you convert the sine function smoothly, you're halfway through understanding complexities in derivatives involving trigonometric identities.

Exponent Rules

The exponent rules, or laws of exponents, are vital mathematical tools that simplify the differentiation process when involved with powers.
They apply to situations where powers or roots of numbers are involved, significantly transforming a function into a form that’s easier to handle. Using our exercise, you turn \( y = \frac{1}{\sin^2(x)} \) into \( y = \sin^{-2}(x) \).
This conversion is guided by the law that states that \( a^{-n} = \frac{1}{a^n} \), allowing expressions to be simplified for differentiation.In general, the key exponent rules include:

• Multiplication: \( x^a \cdot x^b = x^{a+b} \)
• Division: \( \frac{x^a}{x^b} = x^{a-b} \)
• Power of a power: \((x^a)^b = x^{a \cdot b} \)
• Negative exponent: \( x^{-a} = \frac{1}{x^a} \)

Employing these rules transforms and simplifies expressions so that differentiation is smoother and easily applied. They fundamentally support handling any power where direct application of differential calculus would otherwise be cumbersome.