Problem 9 - Calculus: One and Several Variables

Question

Find \(f^{(9)}(0)\) $$f(x)=x^{2} \sin x$$

Step-by-Step Solution

Answer

\(f^{(9)}(0)= 0\)

Step 1: Find the first derivative

Using the product rule, \(d[uv] = udv + vdu\), the first derivative of \(f(x)=x^{2} \sin x\) is \(f'(x) = 2x \sin x + x^{2} \cos x\)

Step 2: Establish the derivative pattern

If we continue finding derivatives, we see a pattern as the derivatives of sine and cosine functions are cyclic. The derivative of a sine function is a cosine function and the derivative of a cosine function is negative sine. This cycle repeats after the fourth derivative. By recognizing this, we can predict what \(f^{(9)}(x)\) will be without finding all the derivatives through to the eighth. The fourth derivative at zero gives us 2, so the eighth derivative at zero gives us another 2. This means the ninth derivative at zero will be the same as the first derivative at zero.

Step 3: Determine the ninth derivative at zero

Following the pattern, \(f^{(9)}(x)\) is the same as \(f'(0)\), which is 0

Key Concepts

Product Rule

When we deal with functions that are the product of two simpler functions, like in the given function \(f(x) = x^{2} \sin x\), we use the product rule to find its derivative. The product rule is vital in differentiation and is stated as:

• If \( u(x) \) and \( v(x) \) are functions of \( x \), the derivative of their product is \((uv)' = u'v + uv'\).

This means you take the derivative of the first function and multiply it by the second, and then add the first function times the derivative of the second. In our exercise, \(u(x) = x^2\) and \(v(x) = \sin x\). So their derivatives are \(u'(x) = 2x\) and \(v'(x) = \cos x\) respectively. By applying the product rule, we find that the first derivative is \(f'(x) = 2x \sin x + x^2 \cos x\). Breaking it down helps us understand how each part is derived before moving to higher derivatives.

Trigonometric Functions

Trigonometric functions like sine and cosine have unique properties, particularly their cyclic derivatives. This cyclic behavior simplifies understanding when taking higher derivatives. Here's what you should know:

• The derivative of \( \sin x \) is \( \cos x \), and the derivative of \( \cos x \) is \(-\sin x \).
• Continuing this pattern: The derivative of \(-\sin x\) is \(-\cos x\), and \(-\cos x\) becomes \( \sin x \) again.
• These derivatives repeat every four functions: \( \sin x, \cos x, -\sin x, -\cos x\).

In our problem, this cyclic nature helps identify the behavior of the function with higher order derivatives. You don't need to compute every derivative separately up to \(f^{(9)}(x)\); recognizing the pattern and cycle allows you to predict accurately.

Derivative Patterns

In calculus, you can detect patterns that help predict higher order derivatives. Recognizing these patterns means not calculating each derivative individually, saving time and effort. Here's how it applies:

• The problem has already recognized a repeating pattern for the function \(x^2 \sin x\) when differentiating multiple times.
• Once established, the cyclic behavior of sine and cosine indicates how the derivatives repeat every four steps.
• By using this cycle, we find that the ninth derivative at a point can relate back to earlier derivatives, significantly helping solve for \(f^{(9)}(0)\) without exhaustive computation.

Therefore, understanding these patterns in derivatives, combined with recognizing how trigonometric functions behave, allows for a structured and simplified approach to problems involving higher order derivatives.