taylor 2.5x^3-3x{2}

Lösung

taylor

2.5 x^{3} - 3 x {2}

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

Schritte zur Lösung

Apply the Taylor Formula

= 0 + \frac{\frac{d}{d x} ( 2.5 x ^{3} - 3 x { 2 } ) ( 0 )}{1 !} x + \frac{\frac{d ^{2}}{d x ^{2}} ( 2.5 x ^{3} - 3 x { 2 } ) ( 0 )}{2 !} x^{2} + \frac{\frac{d ^{3}}{d x ^{3}} ( 2.5 x ^{3} - 3 x { 2 } ) ( 0 )}{3 !} x^{3} + \dots

Evaluate Derivatives

= 0 + \frac{- 6}{1 !} x + \frac{0}{2 !} x^{2} + \frac{15}{3 !} x^{3} + \frac{0}{4 !} x^{4} + \frac{0}{5 !} x^{5} + \frac{0}{6 !} x^{6} + \frac{0}{7 !} x^{7} + \frac{0}{8 !} x^{8} + \frac{0}{9 !} x^{9} + \frac{0}{10 !} x^{10}

Fasse zusammen

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

t an g e n t y = x^{3} - 9 x

\int - \frac{1}{x} d x

\frac{\partial}{\partial x} (x^{2} + y^{3})

\frac{d}{d x} (y - \frac{1}{( x + 3 )})

\frac{d y}{d x} - 7 y = e^{x}