Problem 18 - Calculus Single Variable

Question

An expression for \(f(x)\) is given. Compute the first, second, and third derivatives of \(f(x)\) with respect to \(x\). \(x \sin (x)\)

Step-by-Step Solution

Answer

1st Derivative: \( \sin(x) + x \cos(x) \), 2nd: \( 2\cos(x) - x \sin(x) \), 3rd: \( -3\sin(x) - x\cos(x) \).

Step 1: Understanding the Problem

We need to find the first, second, and third derivatives of the function \(f(x) = x \sin(x)\) with respect to \(x\). This involves applying different rules of differentiation.

Step 2: Calculating the First Derivative

We use the product rule for differentiation which is \( (uv)' = u'v + uv' \) for functions \(u\) and \(v\). Let \(u = x\) and \(v = \sin(x)\). First, find \(u'\) and \(v'\):\[u' = 1, \quad v' = \cos(x)\.\] Apply the product rule:\[f'(x) = (x)' \sin(x) + x \cdot (\sin(x))' = 1 \cdot \sin(x) + x \cdot \cos(x) = \sin(x) + x \cos(x)\.\] Thus, the first derivative is \( f'(x) = \sin(x) + x \cos(x)\).

Step 3: Calculating the Second Derivative

Differentiate \( f'(x) = \sin(x) + x \cos(x) \) using the basic derivative rules:The derivative of \( \sin(x) \) is \( \cos(x)\) and apply the product rule again on \( x \cos(x) \). Let \( u = x \) and \( v = \cos(x) \) with \( u' = 1 \) and \( v' = -\sin(x) \):\[f''(x) = \cos(x) + (x)' \cos(x) + x \cdot (-\sin(x)) = \cos(x) + 1 \cdot \cos(x) - x \sin(x) = 2\cos(x) - x \sin(x)\.\] Hence, the second derivative is \( f''(x) = 2\cos(x) - x \sin(x)\).

Step 4: Calculating the Third Derivative

Differentiate \( f''(x) = 2\cos(x) - x \sin(x) \) using basic differentiation rules: \(2\cos(x)\) differentiates to \(-2\sin(x)\). For \( -x\sin(x) \), use the product rule again. Let \( u = x \) and \( v = -\sin(x) \) where \( u' = 1 \) and \( v' = -\cos(x) \):\[f'''(x) = (-2\sin(x)) + ((x)' \cdot -\sin(x) + x \cdot (-\cos(x))) = -2\sin(x) - \sin(x) - x\cos(x) = -3\sin(x) - x\cos(x)\.\] Thus, the third derivative is \( f'''(x) = -3\sin(x) - x\cos(x)\).

Key Concepts

Product Rule in Differentiation

The product rule is a fundamental tool in calculus used when you need to differentiate a product of two functions. Imagine you have two functions, denoted as \( u(x) \) and \( v(x) \), and you want to find the derivative of their product. The product rule states that the derivative of \( u(x)v(x) \) is given by:

• \( (uv)' = u'v + uv' \)

This means you multiply the derivative of the first function, \( u' \), by the second function, \( v \), and then add it to the product of the first function, \( u \), and the derivative of the second function, \( v' \).

In our exercise, \( u(x) = x \) and \( v(x) = \sin(x) \). Differentiating these gives \( u' = 1 \) and \( v' = \cos(x) \). Substituting back in, we get the first derivative:

• \( f'(x) = (x)' \sin(x) + x(\sin(x))' = \sin(x) + x \cos(x) \)

The product rule helps us understand how the derivatives of individual parts combine when working with products of functions.

Understanding the Second Derivative

The second derivative provides insights into the concavity and the rate of change of the slope of a function. It is simply the derivative of the derivative. In our case, we need to differentiate \( f'(x) = \sin(x) + x\cos(x) \) again.

Using the product rule on \( x \cos(x) \), with \( u(x) = x \) and \( v(x) = \cos(x) \), we find \( u' = 1 \) and \( v' = -\sin(x) \). Now we calculate the second derivative:

• \( f''(x) = \cos(x) + (x' \cdot \cos(x)) + x \cdot (-\sin(x)) = 2\cos(x) - x\sin(x) \)

The sign and value of the second derivative tell us if the function is concave up or down. Here, for different \( x \), \( f''(x) \) will reveal where the slope of \( f(x) \) is increasing or decreasing.

Exploring the Third Derivative

Differentiating further from the second derivative can help us understand the more intricate behaviors of a function's graph. The third derivative is called the jerk in physics, describing the rate of change of acceleration. For calculus, it helps anticipate future curvature changes and inflection tendencies in a function.

Starting with \( f''(x) = 2\cos(x) - x\sin(x) \), differentiate using basic rules and the product rule on \(-x\sin(x)\). Let \( u(x) = x \), \( v(x) = -\sin(x) \), so \( u' = 1 \) and \( v' = -\cos(x) \).

• \( f'''(x) = (-2\sin(x)) + ((x)' \cdot -\sin(x) + x \cdot (-\cos(x))) = -3\sin(x) - x \cos(x) \)

The third derivative will often be more challenging to interpret intuitively. It involves more in-depth analysis to see how these values of \( x \) affect the curvature changes on the graph of the original function.