Question 10 ( 4 points)
Let $g (x) = x \cos (3 x)$ . Find the Maclaurin series of its derivative, $g^{'} (x)$
None of the other answers are correct.
$3 x - \frac{9}{2} x^{3} + \frac{81}{40} x^{5} + \dots$
$3 - \frac{3}{2} x^{2} + \frac{5}{8} x^{4} + \dots$
$1 - \frac{27}{2} x^{2} + \frac{135}{8} x^{4} + \dots$
$3 - \frac{27}{2} x^{2} + \frac{81}{8} x^{4} + \dots$

Answer

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Answer

$g^{'} (x) = 3 - \frac{9}{2} x^{2} + \frac{81}{40} x^{4} + \dots$

Steps

Step 1 ： $g (x) = x \cos (3 x)$

Step 2 ： $g^{'} (x) = \cos (3 x) - 3 x \sin (3 x)$

Step 3 ： $g^{'} (0) = 3, g^{''} (0) = - \frac{9}{2}, g^{'''} (0) = \frac{81}{40}$

Step 4 ： $g^{'} (x) = 3 - \frac{9}{2} x^{2} + \frac{81}{40} x^{4} + \dots$