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To solve this problem, we need to find the Taylor polynomial of degree 6 for the function \( f(x) = x \cos(x^2) \).

### Step 1: Taylor Expansion of \( \cos(x^2) \)
The Taylor series expansion for \( \cos(x) \) around \( x = 0 \) is:

\[
\cos(x) = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!} + \dots
\]

For \( \cos(x^2) \), we substitute \( x^2 \) into the expansion:

\[
\cos(x^2) = 1 - \frac{x^4}{2!} + \frac{x^8}{4!} - \frac{x^{12}}{6!} + \dots
\]

### Step 2: Multiply by \( x \)
Now, multiply this expansion by \( x \) to find \( x \cos(x^2) \):

\[
f(x) = x \left(1 - \frac{x^4}{2!} + \frac{x^8}{4!} - \dots \right)
\]

\[
f(x) = x - \frac{x^5}{2!} + \frac{x^9}{4!} - \dots
\]

Since we only need the polynomial up to degree 6, we stop after the \( x^5 \) term:

\[
f(x) = x - \frac{x^5}{2}
\]

### Step 3: Verify the Options
Looking at the options, the correct Taylor polynomial of degree 6 is the first one:

\[
x - \frac{x^5}{2}
\]

Thus, the correct answer is the first option.

Would you like more details on any part of the process?

### 5 Related Questions:
1. What is the general form of a Taylor series expansion?
2. How do you find the Taylor series for higher-order functions?
3. How does the Taylor series for \( \sin(x) \) differ from that of \( \cos(x) \)?
4. Can the Taylor series approximation be applied to any function?
5. How do you determine the degree of a Taylor polynomial?

### Tip:
When expanding functions using Taylor series, always substitute the argument of the function (e.g., \( x^2 \) instead of \( x \)) carefully and adjust the terms accordingly to avoid mistakes.

Find the Taylor polynomial of degree 6 of the function x cos(x^2).

Learn how to find the Taylor polynomial of degree 6 for the function x cos(x^2), using the expansion of the cosine function and polynomial approximation methods. This step-by-step guide is suitable for undergraduate calculus students.

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