What is taylor 1/(x-1),7 ?

The answer to taylor 1/(x-1),7 is 1/6-1/36 (x-7)+1/216 (x-7)^2-1/1296 (x-7)^3+1/7776 (x-7)^4+\ldots

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taylor 1/(x-1),7

taylor

\frac{1}{x - 1}, 7

\frac{1}{6} - \frac{1}{36} (x - 7) + \frac{1}{216} (x - 7)^{2} - \frac{1}{1296} (x - 7)^{3} + \frac{1}{7776} (x - 7)^{4} + \dots

Solution steps

Apply the Taylor Formula

= \frac{1}{6} + \frac{\frac{d}{d x} ( \frac{1}{x - 1} ) ( 7 )}{1 !} (x - 7) + \frac{\frac{d ^{2}}{d x ^{2}} ( \frac{1}{x - 1} ) ( 7 )}{2 !} (x - 7)^{2} + \frac{\frac{d ^{3}}{d x ^{3}} ( \frac{1}{x - 1} ) ( 7 )}{3 !} (x - 7)^{3} + \dots

Evaluate Derivatives

= \frac{1}{6} + \frac{- \frac{1}{36}}{1 !} (x - 7) + \frac{\frac{1}{108}}{2 !} (x - 7)^{2} + \frac{- \frac{1}{216}}{3 !} (x - 7)^{3} + \frac{\frac{1}{324}}{4 !} (x - 7)^{4} + \dots

Refine

= \frac{1}{6} - \frac{1}{36} (x - 7) + \frac{1}{216} (x - 7)^{2} - \frac{1}{1296} (x - 7)^{3} + \frac{1}{7776} (x - 7)^{4} + \dots

\frac{d}{d x} (3 - 3 sin (x))

\int \frac{1}{u ^{2} + 2 u} d u

\int t^{2} e^{- 21 t} d t

y^{^{'}} = 3 x y^{2}

\frac{d}{d x} (\frac{x}{3 x ^{2} + 4})

What is taylor 1/(x-1),7 ?
The answer to taylor 1/(x-1),7 is 1/6-1/36 (x-7)+1/216 (x-7)^2-1/1296 (x-7)^3+1/7776 (x-7)^4+\ldots