Problem 177 - Calculus Volume 1

Question

For the following exercises, find \(\frac{d y}{d x}\) for the given functions. $$y=x^{2} \cot x$$

Step-by-Step Solution

Answer

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

Step 1: Identify the Function Structure

The given function is a product of two functions: a polynomial function, \(x^2\), and a trigonometric function, \(\cot x\). To differentiate this, we'll use the product rule of differentiation.

Step 2: Recall the Product Rule

The product rule states that if you have a function \(y = u(x) \, v(x)\), then the derivative \(\frac{dy}{dx} = u'(x) \, v(x) + u(x) \, v'(x)\). Here, \(u(x) = x^2\) and \(v(x) = \cot x\).

Step 3: Differentiate \(u(x) = x^2\)

The derivative of \(u(x) = x^2\) with respect to \(x\) is \(u'(x) = 2x\).

Step 4: Differentiate \(v(x) = \cot x\)

The derivative of \(v(x) = \cot x\) with respect to \(x\) is \(v'(x) = -\csc^2 x\), using the derivative rule for \(\cot x\).

Step 5: Apply the Product Rule

Substitute back into the product rule formula to get: \[\frac{dy}{dx} = (2x)(\cot x) + (x^2)(-\csc^2 x)\].

Step 6: Simplify the Expression

Simplify the expression from Step 5: \[\frac{dy}{dx} = 2x \cot x - x^2 \csc^2 x\].

Key Concepts

Product Rule

The product rule is an essential technique in differentiation, especially when you're dealing with the product of two different types of functions. It's a method for finding the derivative of a product of two functions, say \(u(x)\) and \(v(x)\). In essence, if you have \(y = u(x) \cdot v(x)\), the rule states that the derivative \(\frac{dy}{dx}\) can be calculated as follows:

• First, differentiate \(u(x)\) to get \(u'(x)\).
• Then, differentiate \(v(x)\) to get \(v'(x)\).
• Finally, apply the formula: \(u'(x) \cdot v(x) + u(x) \cdot v'(x)\).

This means you multiply the derivative of the first function by the second function and add it to the first function multiplied by the derivative of the second function. It's a simple switch to remember. By breaking down the problem into these clear steps, the product rule helps solve complex differentiation problems seamlessly.

Polynomial Function

Polynomial functions are some of the simpler functions you will encounter in calculus. They are equations that consist of powers of \(x\) with constant coefficients. For example, in our exercise, \(x^2\) is a polynomial function. Polynomials can be easily differentiated using basic rules.

• To differentiate a polynomial like \(x^n\), use the power rule.
• The power rule states that the derivative of \(x^n\) is \(nx^{n-1}\).

Thus, the derivative of \(x^2\) becomes \(2x\). The great thing about polynomials is they make practicing differentiation straightforward. Once you grasp the rules, you'll find that they apply consistently no matter the degree of the polynomial.

Trigonometric Function

Trigonometric functions like sine, cosine, and cotangent play a critical role in calculus because they frequently appear in applications from engineering to physics. In our current problem, the trigonometric function is \(\cot x\).

• The derivative of \(\cot x\), a less common trigonometric function, is \(-\csc^2 x\).
• It follows a standard trigonometric derivative rule people often memorize along with sine and cosine derivatives.

Understanding the derivatives of trigonometric functions not only helps solve problems quickly but also builds a foundation for tackling more advanced topics. It's helpful to memorize or keep a chart of these basic derivatives handy for reference during your studies.

Differentiation

Differentiation is a core concept in calculus. It's the process through which you find the derivative, which shows how a function changes at any given point. The purpose of differentiation is to make sense of the rate of change — how fast or slow a function is increasing or decreasing. In essence, differentiation provides a tool for analyzing the behavior of functions.

• For calculating derivatives, different rules apply based on the kind of function, such as product rule, chain rule, etc.
• In practice, differentiation simplifies to applying these rules step by step.

In our exercise, by differentiating \(y = x^{2} \cot x\), we employ both the product rule and trigonometric differentiation methods. Through careful step-by-step application of these differentiation rules, one can find how \(y\) gradually changes concerning \(x\). As you progress in calculus, mastering these foundational skills will make more complex topics easier to understand.