# Find the Maclaurin series for f(x) = cos(x^{2}) and use it to determine f^{(4)}(0).

Maclaurin series is nothing but Taylor series about the point x = 0.

## Answer: cos (x^{2}) = 1 + x^{4}/2! + x^{8}/4! + x^{12}/6! + x^{16}/8! + ..., the value of f^{(4)}(0) = 12

Let us write the Maclaurin series for f(x) = cos(x^{2}) and use it to determine f^{(4)}(0).

**Explanation:**

Maclaurin series of the function f(x) is given by

f(x) = f(0) + f'(0)x + f"(0) x^{2} / 2! + f'"(0) x^{3} / 3! + ... + f^{(n)}(0) x^{n} / n! + ... ------- (1)

Using the definition of Maclaurin series, we can write cos x as

cos x = 1 + x^{2}/2! + x^{4}/4! + x^{6}/6! + x^{8}/8! + ...

Replace x by x^{2}, we get

cos (x^{2}) = 1 + x^{4}/2! + x^{8}/4! + x^{12}/6! + x^{16}/8! + ...-------- (2)

f^{(4)}(0) is the 4^{th} derivative of f(x) evaluated at x = 0.

Comparing the coefficients of x^{4} from equations (1) and (2), we get

f^{(4)}(0) / 4! = 1/2!

f^{(4)}(0) = 4! / 2! = 12

### So, f^{(4)}(0) = 12

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