What is taylor x^{1/2},1 ?

The answer to taylor x^{1/2},1 is 1+1/2 (x-1)-1/8 (x-1)^2+1/16 (x-1)^3-5/128 (x-1)^4+\ldots

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taylor x^{1/2},1

taylor

x^{\frac{1}{2}}, 1

1 + \frac{1}{2} (x - 1) - \frac{1}{8} (x - 1)^{2} + \frac{1}{16} (x - 1)^{3} - \frac{5}{128} (x - 1)^{4} + \dots

Solution steps

Apply the Taylor Formula

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

Evaluate Derivatives

= 1 + \frac{\frac{1}{2}}{1 !} (x - 1) + \frac{- \frac{1}{4}}{2 !} (x - 1)^{2} + \frac{\frac{3}{8}}{3 !} (x - 1)^{3} + \frac{- \frac{15}{16}}{4 !} (x - 1)^{4} + \dots

Refine

= 1 + \frac{1}{2} (x - 1) - \frac{1}{8} (x - 1)^{2} + \frac{1}{16} (x - 1)^{3} - \frac{5}{128} (x - 1)^{4} + \dots

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

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

\frac{d}{d x} (\frac{4 x}{9 - tan ( x )})

\frac{\partial}{\partial z} (x e^{w (x, z)})

x sin (π x)

What is taylor x^{1/2},1 ?
The answer to taylor x^{1/2},1 is 1+1/2 (x-1)-1/8 (x-1)^2+1/16 (x-1)^3-5/128 (x-1)^4+\ldots