What is taylor 1/x-1 ?

The answer to taylor 1/x-1 is -(x-1)+(x-1)^2-(x-1)^3+(x-1)^4-(x-1)^5+\ldots

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taylor 1/x-1

taylor

\frac{1}{x} - 1

- (x - 1) + (x - 1)^{2} - (x - 1)^{3} + (x - 1)^{4} - (x - 1)^{5} + \dots

Solution steps

Apply the Taylor Formula

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

Evaluate Derivatives

= 0 + \frac{- 1}{1 !} (x - 1) + \frac{2}{2 !} (x - 1)^{2} + \frac{- 6}{3 !} (x - 1)^{3} + \frac{24}{4 !} (x - 1)^{4} + \frac{- 120}{5 !} (x - 1)^{5} + \dots

Refine

= - (x - 1) + (x - 1)^{2} - (x - 1)^{3} + (x - 1)^{4} - (x - 1)^{5} + \dots

\frac{\partial}{\partial x} (2 y - 3 x + 5)

\int (e^{6 x} + e^{- 9 x}) d x

\int x^{2} \cdot cos (n x) d x

\frac{d}{d x} (\frac{2 x}{x - 4})

\frac{d}{d x} (\frac{2 x}{x ^{2} + 9})

What is taylor 1/x-1 ?
The answer to taylor 1/x-1 is -(x-1)+(x-1)^2-(x-1)^3+(x-1)^4-(x-1)^5+\ldots