tangent (x^2+1)^{x+1}

Lösung

tangente von

(x^{2} + 1)^{x + 1}

y = (ln (a_{0}^{2} + 1) + \frac{2 a _{0} ( a _{0} + 1 )}{a _{0}^{2} + 1}) (a_{0}^{2} + 1)^{a_{0} + 1} x + (a_{0}^{2} + 1)^{a_{0} + 1} - (ln (a_{0}^{2} + 1) + \frac{2 a _{0} ( a _{0} + 1 )}{a _{0}^{2} + 1}) (a_{0}^{2} + 1)^{a_{0} + 1} a_{0}

Schritte zur Lösung

Berechne den Graphen der Tangente für den allgemeinen Punkt

x = a_{0}

Finde den Tangentenpunkt:

(a_{0}, (a_{0}^{2} + 1)^{a_{0} + 1})

Finde die Steigung von

y = (x^{2} + 1)^{x + 1} : \frac{d y}{d x} = (ln (x^{2} + 1) + \frac{2 x ( x + 1 )}{x ^{2} + 1}) (x^{2} + 1)^{x + 1}

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 = (ln (a_{0}^{2} + 1) + \frac{2 a _{0} ( a _{0} + 1 )}{a _{0}^{2} + 1}) (a_{0}^{2} + 1)^{a_{0} + 1}

Finde den Graphen mit Steigung m=

(ln (a_{0}^{2} + 1) + \frac{2 a _{0} ( a _{0} + 1 )}{a _{0}^{2} + 1}) (a_{0}^{2} + 1)^{a_{0} + 1}

, der durch den Punkt

(a_{0}, (a_{0}^{2} + 1)^{a_{0} + 1})

verläuft:

y = (ln (a_{0}^{2} + 1) + \frac{2 a _{0} ( a _{0} + 1 )}{a _{0}^{2} + 1}) (a_{0}^{2} + 1)^{a_{0} + 1} x + (a_{0}^{2} + 1)^{a_{0} + 1} - (ln (a_{0}^{2} + 1) + \frac{2 a _{0} ( a _{0} + 1 )}{a _{0}^{2} + 1}) (a_{0}^{2} + 1)^{a_{0} + 1} a_{0}

y = (ln (a_{0}^{2} + 1) + \frac{2 a _{0} ( a _{0} + 1 )}{a _{0}^{2} + 1}) (a_{0}^{2} + 1)^{a_{0} + 1} x + (a_{0}^{2} + 1)^{a_{0} + 1} - (ln (a_{0}^{2} + 1) + \frac{2 a _{0} ( a _{0} + 1 )}{a _{0}^{2} + 1}) (a_{0}^{2} + 1)^{a_{0} + 1} a_{0}

\frac{d}{d x} (x^{3} + 4 x - 15)

d er i v a t i v e (x^{2} - 9 x) 7^{x}

d er i v a t i v e f (x) = (3 x - 2)^{5} (x + 1)^{3}

y - 2 y^{^{'}} + y = 0

\frac{d}{d x} (x^{4 c o s (x)})