What is taylor (x^2)/(1+x),0 ?

The answer to taylor (x^2)/(1+x),0 is x^2-x^3+x^4-x^5+x^6+\ldots

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taylor (x^2)/(1+x),0

taylor

\frac{x ^{2}}{1 + x}, 0

x^{2} - x^{3} + x^{4} - x^{5} + x^{6} + \dots

Solution steps

Apply the Taylor Formula

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

Evaluate Derivatives

= 0 + \frac{0}{1 !} x + \frac{2}{2 !} x^{2} + \frac{- 6}{3 !} x^{3} + \frac{24}{4 !} x^{4} + \frac{- 120}{5 !} x^{5} + \frac{720}{6 !} x^{6} + \dots

Refine

= x^{2} - x^{3} + x^{4} - x^{5} + x^{6} + \dots

8 (sin (t) y^{^{'}} + cos (t) y) = cos (t) sin^{5} (t)

\frac{1}{1 + 5 x ^{2}}

f (x) = 5 x + 1, at x = 7

e^{- 3 x}

f (x) = x^{2} cot (x)

What is taylor (x^2)/(1+x),0 ?
The answer to taylor (x^2)/(1+x),0 is x^2-x^3+x^4-x^5+x^6+\ldots