taylor f(x)=x^2-2x+4

Lösung

taylor

f (x) = x^{2} - 2 x + 4

4 - 2 x + x^{2}

Schritte zur Lösung

Apply the Taylor Formula

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

Evaluate Derivatives

= 4 + \frac{- 2}{1 !} x + \frac{2}{2 !} x^{2} + \frac{0}{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

= 4 - 2 x + x^{2}

\frac{\partial}{\partial x} (x ln (z^{7} + x^{7}))

d er i v a t i v e x (x^{4} + 2)^{3}

d er i v a t i v e sec (θ) tan (θ)

\frac{d}{d x} (lo g_{1} (x))

\int y^{3} + 1 d y