Question

Find the requested derivative. \(f(x)=x \sin x ;\) find \(f^{(4)}(x)\).

Short answer

The fourth derivative of \( f(x) = x \sin x \) is \( f^{(4)}(x) = -x \sin x \).

Step-by-step solution

Identify the function

We are given the function \( f(x) = x \sin x \). To find the fourth derivative \( f^{(4)}(x) \), we'll need to differentiate \( f(x) \) four times.

First Derivative

To find the first derivative \( f'(x) \), apply the product rule: if \( f(x) = u(x) \cdot v(x) \), then \( f'(x) = u'(x) \cdot v(x) + u(x) \cdot v'(x) \). Here, \( u(x) = x \) and \( v(x) = \sin x \). So, \( u'(x) = 1 \) and \( v'(x) = \cos x \). Thus,\[f'(x) = 1 \cdot \sin x + x \cdot \cos x = \sin x + x \cos x\]

Second Derivative

Differentiate \( f'(x) = \sin x + x \cos x \) to get \( f''(x) \). Again, using the sum and product rules:- The derivative of \( \sin x \) is \( \cos x \).- Apply the product rule to \( x\cos x \): - \( u(x) = x \), \( v(x) = \cos x \) - \( u'(x) = 1 \), \( v'(x) = -\sin x \)\[f''(x) = \cos x + (1 \cdot \cos x + x \cdot (-\sin x)) = \cos x + \cos x - x \sin x = 2\cos x - x \sin x\]

Third Derivative

Differentiate \( f''(x) = 2\cos x - x \sin x \) to get \( f'''(x) \). Again, use derivation rules:- The derivative of \( 2\cos x \) is \( -2\sin x \).- Apply product rule to \( -x\sin x \): - \( u(x) = x \), \( v(x) = -\sin x \) - \( u'(x) = 1 \), \( v'(x) = -\cos x \)\[f'''(x) = -2 \sin x - (1 \cdot (-\sin x) + x \cdot (-\cos x)) = -2 \sin x + \sin x + x \cos x = - \sin x + x \cos x\]

Fourth Derivative

Differentiating \( f'''(x) = -\sin x + x\cos x \) to get \( f^{(4)}(x) \):- The derivative of \( -\sin x \) is \( -\cos x \).- Apply product rule for \( x\cos x \): - \( u(x) = x \), \( v(x) = \cos x \) - \( u'(x) = 1 \), \( v'(x) = -\sin x \)\[f^{(4)}(x) = - \cos x + (1 \cdot \cos x + x \cdot (-\sin x)) = - \cos x + \cos x - x \sin x = -x \sin x\]

Key concepts

Product Rule

The concept of the product rule is crucial when dealing with the derivative of a function that is the product of two simpler functions. Imagine you have a function expressed as the product of two functions, like our function \( f(x) = x \sin x \). Here, \( x \) and \( \sin x \) are our two simple functions.

• The product rule states that the derivative of the product of two functions \( u(x) \) and \( v(x) \) is given by: \( f'(x) = u'(x) v(x) + u(x) v'(x) \)
• In a practical scenario, you'll need to differentiate each function separately and then apply the formula. Remember, order and accuracy matter, so structure it step by step.

For instance, in the solution for the first derivative of our function, you have \( u(x) = x \) and \( v(x) = \sin x \). Their derivatives are \( u'(x) = 1 \) and \( v'(x) = \cos x \), respectively. Then applying the product rule, you'll find:\[f'(x) = 1 \cdot \sin x + x \cdot \cos x = \sin x + x \cos x\]This structured approach makes it far easier to tackle derivatives efficiently, especially when handling complex composite functions.

Trigonometric Functions

Trigonometric functions like \( \sin x \), \( \cos x \), and \( \tan x \) often appear in calculus problems due to their periodic nature. Recognizing their properties and derivatives is key to solving these problems efficiently.

• Common derivatives to memorize: \( \frac{d}{dx}(\sin x) = \cos x \) and \( \frac{d}{dx}(\cos x) = -\sin x \).
• These derivatives are essential when applying the product rule and solving higher-order derivatives involving trigonometric functions as seen in \( f(x) = x\sin x \).

Considering \( f(x) = x \sin x \), first we deal with \( \sin x \). In each derivative step, understanding its relationship with \( \cos x \) helps simplify the process. For example, in the third derivative:

• Derivative of \( 2\cos x \) results in \( -2\sin x \).
• Applying this leisurely can prevent calculation mistakes as you move to each derivative step.

Using derivatives of trigonometric functions correctly can make finding solutions much more straightforward in calculus.

Higher-Order Derivatives

Higher-order derivatives involve taking the derivative of a function multiple times. In our exercise, we needed to find \( f^{(4)}(x) \), the fourth derivative of \( f(x) = x \sin x \).

• First Derivative \( f'(x) = \sin x + x\cos x \)
• Second Derivative \( f''(x) = 2\cos x - x\sin x \)
• Third Derivative \( f'''(x) = -\sin x + x\cos x \)
• Fourth Derivative \( f^{(4)}(x) = -x\sin x \)

Higher-order derivatives provide insight into the behavior of a function beyond basic slope analysis. Each subsequent derivative gives a new perspective: - The first derivative \( f'(x) \) offers slope information.- The second derivative \( f''(x) \) might hint at concavity or inflection points.- Further derivatives, like the fourth derivative \( f^{(4)}(x) \), might not relate directly to easily visualized concepts but are critical in deeper analysis.For this reason, understanding the nature of each derivative step is vital. Make sure to rely on past calculations and rules, like the product rule, to move from one order to the next seamlessly.