What is taylor 1/x ,3 ?

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

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taylor 1/x ,3

taylor

\frac{1}{x}, 3

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

Solution steps

Apply the Taylor Formula

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

Evaluate Derivatives

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

Refine

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

\frac{1}{2} x y

6 x + \frac{16}{3}

\frac{\partial}{\partial x} (2 sin (2 x))

5 t

\frac{d}{d x} (\frac{1}{x ^{2} ( 3 x + 1 )})

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