What is taylor 6/(1+36x^2) ?

The answer to taylor 6/(1+36x^2) is 6-216x^2+7776x^4-279936x^6+10077696x^8+\ldots

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taylor 6/(1+36x^2)

taylor

\frac{6}{1 + 36 x ^{2}}

6 - 216 x^{2} + 7776 x^{4} - 279936 x^{6} + 10077696 x^{8} + \dots

Solution steps

Apply the Taylor Formula

= 6 + \frac{\frac{d}{d x} ( \frac{6}{1 + 36 x ^{2}} ) ( 0 )}{1 !} x + \frac{\frac{d ^{2}}{d x ^{2}} ( \frac{6}{1 + 36 x ^{2}} ) ( 0 )}{2 !} x^{2} + \frac{\frac{d ^{3}}{d x ^{3}} ( \frac{6}{1 + 36 x ^{2}} ) ( 0 )}{3 !} x^{3} + \dots

Evaluate Derivatives

= 6 + \frac{0}{1 !} x + \frac{- 432}{2 !} x^{2} + \frac{0}{3 !} x^{3} + \frac{186624}{4 !} x^{4} + \frac{0}{5 !} x^{5} + \frac{- 201553920}{6 !} x^{6} + \frac{0}{7 !} x^{7} + \frac{406332702720}{8 !} x^{8} + \dots

Refine

= 6 - 216 x^{2} + 7776 x^{4} - 279936 x^{6} + 10077696 x^{8} + \dots

\int_{1}^{3} x e^{x^{2}} d x

ln (\frac{3 x ^{4} + 2}{4 x})

\frac{\partial}{\partial x} (y^{3} e^{2 x} + 5 sin (x y))

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

y^{^{''}} + 6 y = 0, y^{^{'}} (1) = 5

What is taylor 6/(1+36x^2) ?
The answer to taylor 6/(1+36x^2) is 6-216x^2+7776x^4-279936x^6+10077696x^8+\ldots