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Beliebt Rechnen >

tangent \sqrt[5]{2x^3+8x}(2.2)

Lösung

tangente von

5 2 x^{3} + 8 x (2.2)

Lösung

y = \frac{0.44 ( 6 a _{0}^{2} + 8 )}{( 2 a _{0}^{3} + 8 a _{0} ) ^{\frac{4}{5}}} x + \frac{220 5 2 a _{0}^{3} + 8 a _{0} ( 2 a _{0}^{3} + 8 a _{0} ) ^{\frac{4}{5}} - 44 a _{0} ( 6 a _{0}^{2} + 8 )}{100 ( 2 a _{0}^{3} + 8 a _{0} ) ^{\frac{4}{5}}}

Schritte zur Lösung

Berechne den Graphen der Tangente für den allgemeinen Punkt

x = a_{0}

Finde den Tangentenpunkt:

(a_{0}, 5 2 a_{0}^{3} + 8 a_{0} \cdot 2.2)

Finde die Steigung von

y = 5 2 x^{3} + 8 x 2.2 : \frac{d y}{d x} = \frac{0.44 ( 6 x ^{2} + 8 )}{( 2 x ^{3} + 8 x ) ^{\frac{4}{5}}}

EN : T i tl e G e n er a l Eq u a t i o n Sl o p e A t P o in t 2 Eq : m = \frac{0.44 ( 6 a _{0}^{2} + 8 )}{( 2 a _{0}^{3} + 8 a _{0} ) ^{\frac{4}{5}}}

Finde den Graphen mit Steigung m=

\frac{0.44 ( 6 a _{0}^{2} + 8 )}{( 2 a _{0}^{3} + 8 a _{0} ) ^{\frac{4}{5}}}

, der durch den Punkt

(a_{0}, 5 2 a_{0}^{3} + 8 a_{0} 2.2)

verläuft:

y = \frac{0.44 ( 6 a _{0}^{2} + 8 )}{( 2 a _{0}^{3} + 8 a _{0} ) ^{\frac{4}{5}}} x + \frac{220 5 2 a _{0}^{3} + 8 a _{0} ( 2 a _{0}^{3} + 8 a _{0} ) ^{\frac{4}{5}} - 44 a _{0} ( 6 a _{0}^{2} + 8 )}{100 ( 2 a _{0}^{3} + 8 a _{0} ) ^{\frac{4}{5}}}

y = \frac{0.44 ( 6 a _{0}^{2} + 8 )}{( 2 a _{0}^{3} + 8 a _{0} ) ^{\frac{4}{5}}} x + \frac{220 5 2 a _{0}^{3} + 8 a _{0} ( 2 a _{0}^{3} + 8 a _{0} ) ^{\frac{4}{5}} - 44 a _{0} ( 6 a _{0}^{2} + 8 )}{100 ( 2 a _{0}^{3} + 8 a _{0} ) ^{\frac{4}{5}}}

Graph

Beliebte Beispiele

limit as n approaches infinity of 5ne^{-n}

n \to \infty lim (5 n e^{- n})

sum from n=1 to infinity of 4/(2^n)

n = 1 \sum \infty \frac{4}{2 ^{n}}

integral of x/(sqrt(1+x^4))

\int \frac{x}{1 + x ^{4}} d x

derivative of 5xy^2

\frac{d}{d x} (5 x y^{2})

integral of 1/(x^2-6x+5)

\int \frac{1}{x ^{2} - 6 x + 5} d x