What is maclaurin f(x)=e^{-x},5 ?

The answer to maclaurin f(x)=e^{-x},5 is 1-x+1/2 x^2-1/6 x^3+1/24 x^4+\ldots

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maclaurin f(x)=e^{-x},5

maclaurin

f (x) = e^{- x}, 5

1 - x + \frac{1}{2} x^{2} - \frac{1}{6} x^{3} + \frac{1}{24} x^{4} + \dots

Solution steps

Apply the Maclaurin Formula

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

Evaluate Derivatives

= 1 + \frac{- 1}{1 !} x + \frac{1}{2 !} x^{2} + \frac{- 1}{3 !} x^{3} + \frac{1}{4 !} x^{4} + \dots

Refine

= 1 - x + \frac{1}{2} x^{2} - \frac{1}{6} x^{3} + \frac{1}{24} x^{4} + \dots

y = ln (5 x \cdot (2 x^{3} - 4 x)^{5})

\int 1 + e^{x} sin (y) d x

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

y^{^{'}} = 3 x

\frac{d}{d x} (e^{- 2 \cdot x})

What is maclaurin f(x)=e^{-x},5 ?
The answer to maclaurin f(x)=e^{-x},5 is 1-x+1/2 x^2-1/6 x^3+1/24 x^4+\ldots