^{2 }+ (f'''(0)/6)*x

^{3}+ (f''''(0)/24)*x

^{4}+ ...

^{2}+ 1/24*x

^{4}+ ...

^{2}/2)*x

^{2}+ (3

^{4}/24)*x

^{4}+ ....

^{4}is 3

^{4}/24, which is 27/8.

Which of the following is the coefficient of x^4 in the Maclaurin series generated by cos (3x)?

a) 27/8

b) 9

c) 1/24

d) 0

e) -27/8

Please help me step by step. Show all your work.

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You know that f(x) can be expanded as a series as follows.

f(x) = f(0) + f'(0)*x + (f''(0)/2)*x^{2 }+ (f'''(0)/6)*x^{3} + (f''''(0)/24)*x^{4} + ...

In case of cos(x), the coefficient of any odd power is zero, which matches well with the fact that cos(x) and even power are even functions.

cos(x) = 1 -1/2*x^{2} + 1/24*x^{4} + ...

When we apply the chain rule in computing derivatives of cos(3x), the Maclaurin series for cos(3x) is computed as follows,

cos(3x) = 1 -(3^{2}/2)*x^{2} + (3^{4}/24)*x^{4} + ....

So, the coefficient of x^{4} is 3^{4}/24, which is 27/8.

The answer: a) 27/8

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