taylor ((e^{2x}-1))/x ,-0.5

Lösung

taylor

\frac{( e ^{2 x} - 1 )}{x}, - 0.5

1.26424 \dots + \frac{- 2 + e}{0.67957 \dots} (x + 0.5) + 0.64241 \dots (x + 0.5)^{2} + 0.30381 \dots (x + 0.5)^{3} + 0.11711 \dots (x + 0.5)^{4} + \dots

Schritte zur Lösung

Apply the Taylor Formula

= 1.26424 \dots + \frac{\frac{d}{d x} ( \frac{e ^{2 x} - 1}{x} ) ( - 0.5 )}{1 !} (x + 0.5) + \frac{\frac{d ^{2}}{d x ^{2}} ( \frac{e ^{2 x} - 1}{x} ) ( - 0.5 )}{2 !} (x + 0.5)^{2} + \frac{\frac{d ^{3}}{d x ^{3}} ( \frac{e ^{2 x} - 1}{x} ) ( - 0.5 )}{3 !} (x + 0.5)^{3} + \dots

Evaluate Derivatives

= 1.26424 \dots + \frac{\frac{- 2 + e}{0.67957 \dots}}{1 !} (x + 0.5) + \frac{1.28482 \dots}{2 !} (x + 0.5)^{2} + \frac{1.82286 \dots}{3 !} (x + 0.5)^{3} + \frac{2.81076 \dots}{4 !} (x + 0.5)^{4} + \dots

Fasse zusammen

= 1.26424 \dots + \frac{- 2 + e}{0.67957 \dots} (x + 0.5) + 0.64241 \dots (x + 0.5)^{2} + 0.30381 \dots (x + 0.5)^{3} + 0.11711 \dots (x + 0.5)^{4} + \dots

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

\frac{\partial}{\partial x} (y x)

(2 x y^{- 1}) d x + (2 - 4 x^{2} y^{- 2}) d y = 0

t an g e n t y = 2 ln (x), at x = 4

\frac{\partial}{\partial x} (x y e^{- x y})