taylor sin(5x)

Lösung

taylor

sin (5 x)

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

Schritte zur Lösung

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

Fasse zusammen

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

\frac{\partial}{\partial x} (x sin (2 y + x))

\frac{d y}{d x} = \frac{1}{4} e^{x} - \frac{y}{x} + \frac{3 x}{4}

\int \frac{4 - 32 x}{1 - 16 x ^{2}} d x

im pl i c i t x^{2} + 4 y^{2} = 4

im pl i c i t \frac{d y}{d x}, y = x^{3}