What is maclaurin (1+x^2)^{1/2} ?

The answer to maclaurin (1+x^2)^{1/2} is 1+1/2 x^2-1/8 x^4+1/16 x^6-5/128 x^8+\ldots

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maclaurin (1+x^2)^{1/2}

maclaurin

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

1 + \frac{1}{2} x^{2} - \frac{1}{8} x^{4} + \frac{1}{16} x^{6} - \frac{5}{128} x^{8} + \dots

Solution steps

Apply the Maclaurin Formula

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

Evaluate Derivatives

= 1 + \frac{0}{1 !} x + \frac{1}{2 !} x^{2} + \frac{0}{3 !} x^{3} + \frac{- 3}{4 !} x^{4} + \frac{0}{5 !} x^{5} + \frac{45}{6 !} x^{6} + \frac{0}{7 !} x^{7} + \frac{- 1575}{8 !} x^{8} + \dots

Refine

= 1 + \frac{1}{2} x^{2} - \frac{1}{8} x^{4} + \frac{1}{16} x^{6} - \frac{5}{128} x^{8} + \dots

\int 3 x^{2} cos (x^{3} + 2) d x

\int 9 e^{0.01 x} d x

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

\frac{d}{d x} (3 x^{2} - 7 x + 10)

f (x) = 2 x^{2} + 5 x - 3

What is maclaurin (1+x^2)^{1/2} ?
The answer to maclaurin (1+x^2)^{1/2} is 1+1/2 x^2-1/8 x^4+1/16 x^6-5/128 x^8+\ldots