What is taylor sin(2x),0 ?

The answer to taylor sin(2x),0 is 2x-4/3 x^3+4/15 x^5-8/315 x^7+4/2835 x^9+\ldots

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taylor sin(2x),0

taylor

sin (2 x), 0

2 x - \frac{4}{3} x^{3} + \frac{4}{15} x^{5} - \frac{8}{315} x^{7} + \frac{4}{2835} x^{9} + \dots

Solution steps

Apply the Taylor Formula

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

Evaluate Derivatives

= 0 + \frac{2}{1 !} x + \frac{0}{2 !} x^{2} + \frac{- 8}{3 !} x^{3} + \frac{0}{4 !} x^{4} + \frac{32}{5 !} x^{5} + \frac{0}{6 !} x^{6} + \frac{- 128}{7 !} x^{7} + \frac{0}{8 !} x^{8} + \frac{512}{9 !} x^{9} + \dots

Refine

= 2 x - \frac{4}{3} x^{3} + \frac{4}{15} x^{5} - \frac{8}{315} x^{7} + \frac{4}{2835} x^{9} + \dots

\frac{d}{d x} (e^{- \frac{x}{6}})

y = x^{2} - 5

\frac{d}{d t} (\frac{e ^{- t}}{1 + t ^{2}})

f (x) = 4 x^{2} + 5 x, at x = - 3

\int \frac{4}{- x ^{2} + 6 x + 5} d x

What is taylor sin(2x),0 ?
The answer to taylor sin(2x),0 is 2x-4/3 x^3+4/15 x^5-8/315 x^7+4/2835 x^9+\ldots