What is taylor sin(2x-1) ?

The answer to taylor sin(2x-1) is -sin(1)+2cos(1)x+2sin(1)x^2-(4cos(1))/3 x^3-(2sin(1))/3 x^4+\ldots

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taylor sin(2x-1)

taylor

sin (2 x - 1)

- sin (1) + 2 cos (1) x + 2 sin (1) x^{2} - \frac{4 cos ( 1 )}{3} x^{3} - \frac{2 sin ( 1 )}{3} x^{4} + \dots

Solution steps

Apply the Taylor Formula

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

Evaluate Derivatives

= - sin (1) + \frac{2 cos ( 1 )}{1 !} x + \frac{4 sin ( 1 )}{2 !} x^{2} + \frac{- 8 cos ( 1 )}{3 !} x^{3} + \frac{- 16 sin ( 1 )}{4 !} x^{4} + \dots

Refine

= - sin (1) + 2 cos (1) x + 2 sin (1) x^{2} - \frac{4 cos ( 1 )}{3} x^{3} - \frac{2 sin ( 1 )}{3} x^{4} + \dots

f (x) = arccot (x)

y = - 2 - 7 x^{2}, (3, - 65)

\frac{d}{d t} (t^{- 1})

(10 + 9 x)^{4} \frac{d y}{d x} = (6 + y)^{2}

x \to - \infty lim (x^{8} e^{x})

What is taylor sin(2x-1) ?
The answer to taylor sin(2x-1) is -sin(1)+2cos(1)x+2sin(1)x^2-(4cos(1))/3 x^3-(2sin(1))/3 x^4+\ldots