What is taylor 1/(x+2),1 ?

The answer to taylor 1/(x+2),1 is 1/3-1/9 (x-1)+1/27 (x-1)^2-1/81 (x-1)^3+1/243 (x-1)^4+\ldots

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taylor 1/(x+2),1

taylor

\frac{1}{x + 2}, 1

\frac{1}{3} - \frac{1}{9} (x - 1) + \frac{1}{27} (x - 1)^{2} - \frac{1}{81} (x - 1)^{3} + \frac{1}{243} (x - 1)^{4} + \dots

Solution steps

Apply the Taylor Formula

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

Evaluate Derivatives

= \frac{1}{3} + \frac{- \frac{1}{9}}{1 !} (x - 1) + \frac{\frac{2}{27}}{2 !} (x - 1)^{2} + \frac{- \frac{2}{27}}{3 !} (x - 1)^{3} + \frac{\frac{8}{81}}{4 !} (x - 1)^{4} + \dots

Refine

= \frac{1}{3} - \frac{1}{9} (x - 1) + \frac{1}{27} (x - 1)^{2} - \frac{1}{81} (x - 1)^{3} + \frac{1}{243} (x - 1)^{4} + \dots

\frac{d y}{d x} = \frac{( 1 + x )}{x y ^{10}}

\frac{\partial}{\partial y} (x e^{- 2 x y})

\int x 2 x - 1 d x

\int \frac{4 x}{3 x ^{2}} d x

d er i v a t i v e f (x) = \frac{x}{x + 7}

What is taylor 1/(x+2),1 ?
The answer to taylor 1/(x+2),1 is 1/3-1/9 (x-1)+1/27 (x-1)^2-1/81 (x-1)^3+1/243 (x-1)^4+\ldots