What is taylor sin(5x),x=0 ?

The answer to taylor sin(5x),x=0 is 5x-125/6 x^3+625/24 x^5-15625/1008 x^7+390625/72576 x^9+\ldots

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taylor sin(5x),x=0

taylor

sin (5 x), x = 0

5 x - \frac{125}{6} x^{3} + \frac{625}{24} x^{5} - \frac{15625}{1008} x^{7} + \frac{390625}{72576} x^{9} + \dots

Solution steps

Apply the Taylor Formula

= 0 + \frac{\frac{d}{d x} ( sin ( 5 x ) ) ( 0 )}{1 !} x + \frac{\frac{d ^{2}}{d x ^{2}} ( sin ( 5 x ) ) ( 0 )}{2 !} x^{2} + \frac{\frac{d ^{3}}{d x ^{3}} ( sin ( 5 x ) ) ( 0 )}{3 !} x^{3} + \dots

Evaluate Derivatives

= 0 + \frac{5}{1 !} x + \frac{0}{2 !} x^{2} + \frac{- 125}{3 !} x^{3} + \frac{0}{4 !} x^{4} + \frac{3125}{5 !} x^{5} + \frac{0}{6 !} x^{6} + \frac{- 78125}{7 !} x^{7} + \frac{0}{8 !} x^{8} + \frac{1953125}{9 !} x^{9} + \dots

Refine

= 5 x - \frac{125}{6} x^{3} + \frac{625}{24} x^{5} - \frac{15625}{1008} x^{7} + \frac{390625}{72576} x^{9} + \dots

\frac{1}{ln ( x + 1 )}

\frac{\partial}{\partial y} (5 x + y)

\frac{d}{d a} (sin^{4} (a) + cos^{4} (a))

\int \frac{sec ( x ) tan ( x )}{- 5 + sec ( x )} d x

x \to 0 + lim (9 cos (9 x))

What is taylor sin(5x),x=0 ?
The answer to taylor sin(5x),x=0 is 5x-125/6 x^3+625/24 x^5-15625/1008 x^7+390625/72576 x^9+\ldots