What is maclaurin e^x+e^{-x} ?

The answer to maclaurin e^x+e^{-x} is 2+x^2+1/12 x^4+1/360 x^6+1/20160 x^8+\ldots

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maclaurin e^x+e^{-x}

maclaurin

e^{x} + e^{- x}

2 + x^{2} + \frac{1}{12} x^{4} + \frac{1}{360} x^{6} + \frac{1}{20160} x^{8} + \dots

Solution steps

Apply the Maclaurin Formula

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

Evaluate Derivatives

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

Refine

= 2 + x^{2} + \frac{1}{12} x^{4} + \frac{1}{360} x^{6} + \frac{1}{20160} x^{8} + \dots

\frac{\partial}{\partial x} (cos (x + e^{y}))

\int \frac{9 x}{x - 1} d x

y = 3 x^{2} - 12 x + 1, (0, 1)

f (x) = x^{2} + 2

\frac{d}{d x} (\frac{x ^{2} - 3}{x + 1})

What is maclaurin e^x+e^{-x} ?
The answer to maclaurin e^x+e^{-x} is 2+x^2+1/12 x^4+1/360 x^6+1/20160 x^8+\ldots