What is taylor 1/(x-1),9 ?

The answer to taylor 1/(x-1),9 is 1/8-1/64 (x-9)+1/512 (x-9)^2-1/4096 (x-9)^3+1/32768 (x-9)^4+\ldots

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taylor 1/(x-1),9

taylor

\frac{1}{x - 1}, 9

\frac{1}{8} - \frac{1}{64} (x - 9) + \frac{1}{512} (x - 9)^{2} - \frac{1}{4096} (x - 9)^{3} + \frac{1}{32768} (x - 9)^{4} + \dots

Solution steps

Apply the Taylor Formula

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

Evaluate Derivatives

= \frac{1}{8} + \frac{- \frac{1}{64}}{1 !} (x - 9) + \frac{\frac{1}{256}}{2 !} (x - 9)^{2} + \frac{- \frac{3}{2048}}{3 !} (x - 9)^{3} + \frac{\frac{3}{4096}}{4 !} (x - 9)^{4} + \dots

Refine

= \frac{1}{8} - \frac{1}{64} (x - 9) + \frac{1}{512} (x - 9)^{2} - \frac{1}{4096} (x - 9)^{3} + \frac{1}{32768} (x - 9)^{4} + \dots

\int \frac{tan ( x )}{cos ^{2} ( x )} d x

x \to \frac{π}{2} lim (7 sin (x))

\int \frac{1}{x ^{- 1} + x ^{- 3}} d x

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

y = - 2 x^{3}, (2, - 16)

What is taylor 1/(x-1),9 ?
The answer to taylor 1/(x-1),9 is 1/8-1/64 (x-9)+1/512 (x-9)^2-1/4096 (x-9)^3+1/32768 (x-9)^4+\ldots