taylor 2^x,1

Lösung

taylor

2^{x}, 1

2 + 2 ln (2) (x - 1) + ln^{2} (2) (x - 1)^{2} + \frac{ln ^{3} ( 2 )}{3} (x - 1)^{3} + \frac{ln ^{4} ( 2 )}{12} (x - 1)^{4} + \dots

Schritte zur Lösung

Apply the Taylor Formula

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

Evaluate Derivatives

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

Fasse zusammen

= 2 + 2 ln (2) (x - 1) + ln^{2} (2) (x - 1)^{2} + \frac{ln ^{3} ( 2 )}{3} (x - 1)^{3} + \frac{ln ^{4} ( 2 )}{12} (x - 1)^{4} + \dots

\int \frac{1}{θ ^{2}} sin (\frac{1}{θ}) cos (\frac{1}{θ}) d θ

\frac{d}{d t} (e^{5 t})

d er i v a t i v e y = e^{9 x}

\frac{d}{d t} (e^{t} cos (2 t))

d er i v a t i v e x^{2} ln (x^{2})