What is taylor 1/(5+7x) ?

The answer to taylor 1/(5+7x) is 1/5-7/25 x+49/125 x^2-343/625 x^3+2401/3125 x^4+\ldots

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taylor 1/(5+7x)

taylor

\frac{1}{5 + 7 x}

\frac{1}{5} - \frac{7}{25} x + \frac{49}{125} x^{2} - \frac{343}{625} x^{3} + \frac{2401}{3125} x^{4} + \dots

Solution steps

Apply the Taylor Formula

= \frac{1}{5} + \frac{\frac{d}{d x} ( \frac{1}{5 + 7 x} ) ( 0 )}{1 !} x + \frac{\frac{d ^{2}}{d x ^{2}} ( \frac{1}{5 + 7 x} ) ( 0 )}{2 !} x^{2} + \frac{\frac{d ^{3}}{d x ^{3}} ( \frac{1}{5 + 7 x} ) ( 0 )}{3 !} x^{3} + \dots

Evaluate Derivatives

= \frac{1}{5} + \frac{- \frac{7}{25}}{1 !} x + \frac{\frac{98}{125}}{2 !} x^{2} + \frac{- \frac{2058}{625}}{3 !} x^{3} + \frac{\frac{57624}{3125}}{4 !} x^{4} + \dots

Refine

= \frac{1}{5} - \frac{7}{25} x + \frac{49}{125} x^{2} - \frac{343}{625} x^{3} + \frac{2401}{3125} x^{4} + \dots

\int \frac{cos ^{2} ( x )}{4 + sin ( 2 x )} d x

\frac{d}{d x} (2 x^{3} - 9 x^{2} + 12 x)

\int_{1}^{2} \frac{x ^{2} + 1}{x} d x

t y^{^{'}} + t y = 1 - y

\int 20 t d t

What is taylor 1/(5+7x) ?
The answer to taylor 1/(5+7x) is 1/5-7/25 x+49/125 x^2-343/625 x^3+2401/3125 x^4+\ldots