maclaurin (x^2+4)sin(5x)

Lösung

maclaurin

(x^{2} + 4) sin (5 x)

20 x - \frac{235}{3} x^{3} + \frac{250}{3} x^{5} - \frac{18125}{504} x^{7} + \frac{15625}{2592} x^{9} + \dots

Schritte zur Lösung

Apply the Maclaurin Formula

= 0 + \frac{\frac{d}{d x} ( ( x ^{2} + 4 ) sin ( 5 x ) ) ( 0 )}{1 !} x + \frac{\frac{d ^{2}}{d x ^{2}} ( ( x ^{2} + 4 ) sin ( 5 x ) ) ( 0 )}{2 !} x^{2} + \frac{\frac{d ^{3}}{d x ^{3}} ( ( x ^{2} + 4 ) sin ( 5 x ) ) ( 0 )}{3 !} x^{3} + \dots

Evaluate Derivatives

= 0 + \frac{20}{1 !} x + \frac{0}{2 !} x^{2} + \frac{- 470}{3 !} x^{3} + \frac{0}{4 !} x^{4} + \frac{10000}{5 !} x^{5} + \frac{0}{6 !} x^{6} + \frac{- 181250}{7 !} x^{7} + \frac{0}{8 !} x^{8} + \frac{2187500}{9 !} x^{9} + \dots

Fasse zusammen

= 20 x - \frac{235}{3} x^{3} + \frac{250}{3} x^{5} - \frac{18125}{504} x^{7} + \frac{15625}{2592} x^{9} + \dots

θ \to 0 + lim (\frac{s}{d})

x \to 1 + lim (x^{\frac{5}{1 - x}})

(y + 1) d x = 2 x y d y

d er i v a t i v e y = (4 x - 3)^{\frac{1}{2}}

x \to \infty lim (2 - 2)