derivative h(t)=(t+1)^{2/3}(2t^2-5)^3

Lösung

ableitung von

h (t) = (t + 1)^{\frac{2}{3}} (2 t^{2} - 5)^{3}

\frac{2 ( 2 t ^{2} - 5 ) ^{3} + 36 t ( 2 t ^{2} - 5 ) ^{2} ( t + 1 )}{3 ( t + 1 ) ^{\frac{1}{3}}}

Schritte zur Lösung

\frac{d}{d t} ((t + 1)^{\frac{2}{3}} (2 t^{2} - 5)^{3})

Wende die Produktregel an:

(f \cdot g)^{'} = f^{'} \cdot g + f \cdot g^{'}

f = (t + 1)^{\frac{2}{3}}, g = (2 t^{2} - 5)^{3}

= \frac{d}{d t} ((t + 1)^{\frac{2}{3}}) (2 t^{2} - 5)^{3} + \frac{d}{d t} ((2 t^{2} - 5)^{3}) (t + 1)^{\frac{2}{3}}

\frac{d}{d t} ((t + 1)^{\frac{2}{3}}) = \frac{2}{3 ( t + 1 ) ^{\frac{1}{3}}}

\frac{d}{d t} ((2 t^{2} - 5)^{3}) = 12 t (2 t^{2} - 5)^{2}

= \frac{2}{3 ( t + 1 ) ^{\frac{1}{3}}} (2 t^{2} - 5)^{3} + 12 t (2 t^{2} - 5)^{2} (t + 1)^{\frac{2}{3}}

Vereinfache

\frac{2}{3 ( t + 1 ) ^{\frac{1}{3}}} (2 t^{2} - 5)^{3} + 12 t (2 t^{2} - 5)^{2} (t + 1)^{\frac{2}{3}} : \frac{2 ( 2 t ^{2} - 5 ) ^{3} + 36 t ( 2 t ^{2} - 5 ) ^{2} ( t + 1 )}{3 ( t + 1 ) ^{\frac{1}{3}}}

= \frac{2 ( 2 t ^{2} - 5 ) ^{3} + 36 t ( 2 t ^{2} - 5 ) ^{2} ( t + 1 )}{3 ( t + 1 ) ^{\frac{1}{3}}}

t an g e n t f (x) = \frac{7 x}{3 + x ^{2}}, at x = 2

x \to 1 lim (- \frac{1}{( x - 1 ) ^{2}})

t an g e n t y = \frac{- 6 x}{x ^{2} + 1}, (0, 0)

d er i v a t i v e f (x) = 4 x^{2} - x

\int \frac{1}{9 x} d x