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

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

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taylor 1/(2x),1

taylor

\frac{1}{2 x}, 1

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

Solution steps

Apply the Taylor Formula

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

Evaluate Derivatives

= \frac{1}{2} + \frac{- \frac{1}{2}}{1 !} (x - 1) + \frac{1}{2 !} (x - 1)^{2} + \frac{- 3}{3 !} (x - 1)^{3} + \frac{12}{4 !} (x - 1)^{4} + \dots

Refine

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

\frac{d}{d x} (cos ((ln (x + 1))^{2} - e^{x^{2}}))

\frac{\partial}{\partial x} (2 x^{\frac{1}{2}} y^{\frac{1}{4}})

x^{3} - \frac{x ^{2}}{40}

y^{^{'}} = 4 + 2 sin (\frac{x}{4}) - \frac{y}{100}

\int \frac{1}{n ( ln ( n ) ) ^{2}}

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