Problem 195 - Calculus Volume 1

Question

Find \(\frac{d^{2} y}{d x^{2}}\) for the given functions. $$ y=2 \csc x $$

Step-by-Step Solution

Answer

\( \frac{d^{2} y}{d x^{2}} = 2 \csc^3 x \)

Step 1: Differentiate to find the first derivative

The given function is \( y = 2 \csc x \). Start by differentiating it with respect to \( x \). We know that the derivative of \( \csc x \) is \(-\csc x \cot x \). Thus:\[ \frac{dy}{dx} = 2 \cdot (-\csc x \cot x) = -2 \csc x \cot x \]

Step 2: Differentiate the first derivative to find the second derivative

Use the product rule to differentiate \( -2 \csc x \cot x \). Let \( u = \csc x \) and \( v = \cot x \), so \( \frac{d}{dx}[uv] = u'v + uv' \). First, compute the derivatives:- \( u' = -\csc x \cot x \)- \( v' = -\csc^2 x \)Now apply the product rule:\[ \frac{d}{dx}[-2 \csc x \cot x] = -2(\csc x \cdot (-\csc^2 x) + (-\csc x \cot x) \cdot \cot x) \]Simplify the expression:\[ = -2 (-\csc^3 x - \csc x \cot^2 x) \]\[ = 2 \csc x (\csc^2 x + \cot^2 x) \]

Step 3: Simplify the second derivative

We know that \( \csc^2 x + \cot^2 x = \csc^2 x (1 + \cot^2 x / \csc^2 x) = \csc^2 x \). This gives:\[ 2 \csc x \cdot \csc^2 x = 2 \csc^3 x \]Thus, the second derivative simplifies to:\[ \frac{d^2 y}{dx^2} = 2 \csc^3 x \]

Key Concepts

Product Rule

When it comes to finding the derivative of the product of two functions, the Product Rule is indispensable. If you have two differentiable functions, say \( u(x) \) and \( v(x) \), and you're tasked with differentiating their product \( u(x)v(x) \), the Product Rule states that:

• \( \frac{d}{dx}[u(x)v(x)] = u'(x)v(x) + u(x)v'(x) \)

The rule essentially tells us to take the derivative of the first function, \( u(x) \), and multiply it by the second function, \( v(x) \), as it is. Then, we add the product of the first function, \( u(x) \), and the derivative of the second function, \( v(x) \).
In the example given, where \( y = -2 \csc x \cot x \), we applied the Product Rule by setting \( u = \csc x \) and \( v = \cot x \). This allowed us to efficiently differentiate the product and simplify it further. Understanding the Product Rule provides leverage when dealing with complexities of multiplied functions under differentiation.

Trigonometric Functions

Trigonometric functions like \( \csc x \), \( \sin x \), and \( \cos x \) often show up in calculus, needing particular attention when it comes to differentiation. These functions are not just about right triangles but also play a crucial role in addressing real-world periodic phenomena.
For the function \( y = 2 \csc x \), it's essential to know how to deal with \( \csc x \), which is the reciprocal of \( \sin x \). It can be defined as:

• \( \csc x = \frac{1}{\sin x} \)

Furthermore, knowing the derivatives of the basic trigonometric functions is important. For example, the derivative of \( \csc x \) is \(-\csc x \cot x\), and for \( \cot x \), it’s \(-\csc^2 x \). These derivations come naturally when we apply the rules of differentiation and can help solve more complex trigonometric expressions effectively.
Familiarity with these functions and how they derive is invaluable, offering the strategic tools needed to tackle further differentiation levels.

Differentiation Techniques

Differentiation involves various techniques, and mastering these can help simplify and solve more complicated problems. Techniques like the Product Rule, Chain Rule, and Knowledge of Derivatives are fundamental.
In our example, we combined an understanding of trigonometric derivatives with the Product Rule to find the second derivative of the given function. Here are some steps to keep in mind:

• Identify the right rule or combination of rules (such as Product or Chain Rule).
• Break down the components of the function and their derivatives.
• Simplify the resulting expression after applying the rules to ascertain the final answer.

These steps are essential when dealing with functions that aren't straightforward. By practicing these techniques, you steadily improve the ability to handle various types of differentiable equations.
Such skill development is crucial as it aids in delving deeper into much more complex calculus problems, ensuring a solid foundation in mathematical analysis.