What is maclaurin sqrt(1+x) ?

The answer to maclaurin sqrt(1+x) is 1+1/2 x-1/8 x^2+1/16 x^3-5/128 x^4+\ldots

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maclaurin sqrt(1+x)

maclaurin

1 + x

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

Solution steps

Apply the Maclaurin Formula

= 1 + \frac{\frac{d}{d x} ( 1 + x ) ( 0 )}{1 !} x + \frac{\frac{d ^{2}}{d x ^{2}} ( 1 + x ) ( 0 )}{2 !} x^{2} + \frac{\frac{d ^{3}}{d x ^{3}} ( 1 + x ) ( 0 )}{3 !} x^{3} + \dots

Evaluate Derivatives

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

Refine

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

\int x^{3} - 9 x d x

\frac{d}{d x} (sin (x + 3) + (- cos (x - \frac{1}{2})))

\int_{- 2}^{- 1} 6 x^{- 3} d x

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

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

What is maclaurin sqrt(1+x) ?
The answer to maclaurin sqrt(1+x) is 1+1/2 x-1/8 x^2+1/16 x^3-5/128 x^4+\ldots