What is maclaurin xcos(4x) ?

The answer to maclaurin xcos(4x) is x-8x^3+32/3 x^5-256/45 x^7+512/315 x^9+\ldots

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maclaurin xcos(4x)

maclaurin

x cos (4 x)

x - 8 x^{3} + \frac{32}{3} x^{5} - \frac{256}{45} x^{7} + \frac{512}{315} x^{9} + \dots

Solution steps

Apply the Maclaurin Formula

= 0 + \frac{\frac{d}{d x} ( x cos ( 4 x ) ) ( 0 )}{1 !} x + \frac{\frac{d ^{2}}{d x ^{2}} ( x cos ( 4 x ) ) ( 0 )}{2 !} x^{2} + \frac{\frac{d ^{3}}{d x ^{3}} ( x cos ( 4 x ) ) ( 0 )}{3 !} x^{3} + \dots

Evaluate Derivatives

= 0 + \frac{1}{1 !} x + \frac{0}{2 !} x^{2} + \frac{- 48}{3 !} x^{3} + \frac{0}{4 !} x^{4} + \frac{1280}{5 !} x^{5} + \frac{0}{6 !} x^{6} + \frac{- 28672}{7 !} x^{7} + \frac{0}{8 !} x^{8} + \frac{589824}{9 !} x^{9} + \dots

Refine

= x - 8 x^{3} + \frac{32}{3} x^{5} - \frac{256}{45} x^{7} + \frac{512}{315} x^{9} + \dots

\frac{s - 5}{s + 1}

\int t^{2} sin (5 t) d t

\frac{d}{d n} ((1 + i)^{n} + (1 - i)^{n})

\frac{d}{d x} (\frac{4}{x} - \frac{1}{x ^{3}})

\int_{1}^{3} f (2 x) d x

What is maclaurin xcos(4x) ?
The answer to maclaurin xcos(4x) is x-8x^3+32/3 x^5-256/45 x^7+512/315 x^9+\ldots