Problem 8 - Essential Calculus

Question

Determine the open intervals on which the graph is concave upward or concave downward. \(y=x+\frac{2}{\sin x}, \quad(-\pi, \pi)\)

Step-by-Step Solution

Answer

The graph of the function \(y = x + \frac{2}{\sin x}\) is concave upward on the interval (-π, 0) and concave downward on the interval (0, π).

Step 1: Find the first derivative of the function

Apply the quotient rule and the chain rule to differentiate \(y = x + \frac{2}{\sin x}\). The derivative should be \(y' = 1 - 2\csc^2 x\).

Step 2: Find the second derivative

The second derivative is the derivative of the first derivative. Differentiating \(y' = 1 - 2\csc^2 x\) gives \(y'' = 4\csc^2 x \cot x\).

Step 3: Determine where the second derivative is positive or negative

Set \(y''\) to be greater or less than zero. This will lead to the solutions for the intervals where the graph is concave up or down. From \(y'' = 4\csc^2 x \cot x\), we can see that \(y''\) is undefined at \( x = 0 \). Also, \(y''\) changes sign at \( x = 0 \). Therefore, \(y'' > 0\) for \(x\) in \((-π, 0)\) and \(y'' < 0\) for \(x\) in \((0, π)\).

Key Concepts

First Derivative

To analyze concavity, we begin by finding the first derivative of the function. The first derivative helps us understand the slope of the function and how it changes across different intervals. For the function \( y = x + \frac{2}{\sin x} \), we need to apply the quotient and chain rules to differentiate it effectively.
The resulting first derivative is \( y' = 1 - 2\csc^2 x \).

• The term \( 1 \) is the derivative of \( x \), which is straightforward.
• For the term \( \frac{2}{\sin x} \), the quotient rule helps us evaluate the change in slope.

With \( y' \), we have a clear picture of how steep the graph is at any \( x \) within the interval \((-\pi, \pi)\). Understanding the behaviour of \( y' \) sets the stage for evaluating concavity.

Second Derivative

The second derivative provides insight into the concavity of the function by telling us whether the graph is curving upwards or downwards. To find it, take the derivative of the first derivative. For \( y' = 1 - 2\csc^2 x \), the second derivative is calculated to be \( y'' = 4\csc^2 x \cot x \).

• If the second derivative, \( y'' \), is positive at a point, the function is concave upward there.
• If \( y'' \) is negative, the function is concave downward.
• Points where \( y'' = 0 \) or is undefined deserve careful attention, as these may indicate inflection points.

When studying \( y'' \), it is important to note where its sign changes and where it might become undefined, helping us identify critical points in the overall graph behavior.

Quotient Rule

The quotient rule is important when differentiating ratios of two functions. It ensures accurate computation of derivatives where one term is in the numerator and another in the denominator.
For a function \( \frac{u}{v} \), where both \( u \) and \( v \) are differentiable functions, the quotient rule is expressed as:\[ \left( \frac{u}{v} \right)' = \frac{u'v - uv'}{v^2} \]
In the exercise, for \( y = x + \frac{2}{\sin x} \), we consider \( u = 2 \) and \( v = \sin x \) when using the quotient rule to find \( y' \).

• \( u' = 0 \) since the derivative of a constant is zero.
• \( v' = \cos x \), the derivative of \( \sin x \).
• Substitute these into the formula to determine the contribution to \( y' \).

This careful step ensures that we can precisely evaluate how the function \( \frac{2}{\sin x} \) evolves as \( x \) changes, aiding in understanding overall slope changes.

Chain Rule

The chain rule is essential when working with composite functions, where one function exists within another. It simplifies the process of differentiation by allowing you to tackle each function layer by layer.
For a composite function \( f(g(x)) \), the chain rule is expressed as:\[ \frac{d}{dx}f(g(x)) = f'(g(x)) \cdot g'(x) \]
In the context of this exercise, the chain rule comes into play when differentiating terms that involve trigonometric functions like \( \csc x \).

• For a function like \( \csc^2 x \), first express it in terms of \( \sin x \).
• Apply the derivative \( (\sin x)' = \cos x \).
• Multiply by the derivative of the outer function \( (\csc^2 x)' \), specifically focusing on the inner workings of this expression.

The chain rule's systematic approach allows us to handle these nested differentiations cleanly, ensuring that calculated derivatives accurately reflect changes in the function's graph.