taylor 1.5^x

Lösung

taylor

1. 5^{x}

1 + 0.40546 \dots x + 0.08220 \dots x^{2} + 0.01110 \dots x^{3} + 0.00112 \dots x^{4} + \dots

Schritte zur Lösung

Apply the Taylor Formula

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

Evaluate Derivatives

= 1 + \frac{0.40546 \dots}{1 !} x + \frac{0.16440 \dots}{2 !} x^{2} + \frac{0.06665 \dots}{3 !} x^{3} + \frac{0.02702 \dots}{4 !} x^{4} + \dots

Fasse zusammen

= 1 + 0.40546 \dots x + 0.08220 \dots x^{2} + 0.01110 \dots x^{3} + 0.00112 \dots x^{4} + \dots

\int \frac{x}{9 - x ^{2}} d x

\int \frac{1}{sin ^{2} ( 3 x )} d x

\frac{d y}{d x} = \frac{sec ( y )}{( x - 3 ) ^{2}}

\frac{d y}{d x} - 2 x y = x y^{2}

d er i v a t i v e \frac{7 - sec ( x )}{tan ( x )}