What is taylor sin(3x),0 ?

The answer to taylor sin(3x),0 is 3x-9/2 x^3+81/40 x^5-243/560 x^7+243/4480 x^9+\ldots

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taylor sin(3x),0

taylor

sin (3 x), 0

3 x - \frac{9}{2} x^{3} + \frac{81}{40} x^{5} - \frac{243}{560} x^{7} + \frac{243}{4480} x^{9} + \dots

Solution steps

Apply the Taylor Formula

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

Evaluate Derivatives

= 0 + \frac{3}{1 !} x + \frac{0}{2 !} x^{2} + \frac{- 27}{3 !} x^{3} + \frac{0}{4 !} x^{4} + \frac{243}{5 !} x^{5} + \frac{0}{6 !} x^{6} + \frac{- 2187}{7 !} x^{7} + \frac{0}{8 !} x^{8} + \frac{19683}{9 !} x^{9} + \dots

Refine

= 3 x - \frac{9}{2} x^{3} + \frac{81}{40} x^{5} - \frac{243}{560} x^{7} + \frac{243}{4480} x^{9} + \dots

12 (t + 1) (\frac{d y}{d t}) - 8 y = 32 t, y (0) = 2

y = \frac{1}{t ^{3} + 2 t ^{2} - 1}

x = - 5, x = 1, y = 3 x, y = x^{2} - 4

\frac{d}{d x} (\frac{4 x - 7}{2 x + 1})

\int \frac{6 x ^{2} - 5 x + 3}{x} d x

What is taylor sin(3x),0 ?
The answer to taylor sin(3x),0 is 3x-9/2 x^3+81/40 x^5-243/560 x^7+243/4480 x^9+\ldots