What is maclaurin 4cos(-x) ?

The answer to maclaurin 4cos(-x) is 4-2x^2+1/6 x^4-1/180 x^6+1/10080 x^8+\ldots

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maclaurin 4cos(-x)

maclaurin

4 cos (- x)

4 - 2 x^{2} + \frac{1}{6} x^{4} - \frac{1}{180} x^{6} + \frac{1}{10080} x^{8} + \dots

Solution steps

Apply the Maclaurin Formula

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

Evaluate Derivatives

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

Refine

= 4 - 2 x^{2} + \frac{1}{6} x^{4} - \frac{1}{180} x^{6} + \frac{1}{10080} x^{8} + \dots

\frac{3}{2} x, 6 x - 2 x, [0, 6]

\frac{d y}{d x} = x (3 - y)

\frac{d}{d x} (\frac{x}{( m + n x )})

\frac{\partial}{\partial x} (x^{2} + 2 x y + 2 y^{2} - 6 x + 10 y + 2)

\frac{d}{d x} (ln (x^{2} 1 - x^{2}))

What is maclaurin 4cos(-x) ?
The answer to maclaurin 4cos(-x) is 4-2x^2+1/6 x^4-1/180 x^6+1/10080 x^8+\ldots