derivative g(u)=((u^3-1)/(u^3+1))^8

Lösung

ableitung von

g (u) = (\frac{u ^{3} - 1}{u ^{3} + 1})^{8}

\frac{48 u ^{2} ( u ^{3} - 1 ) ^{7}}{( u ^{3} + 1 ) ^{9}}

Schritte zur Lösung

\frac{d}{d u} ((\frac{u ^{3} - 1}{u ^{3} + 1})^{8})

Wende die Kettenregel an:

8 (\frac{u ^{3} - 1}{u ^{3} + 1})^{7} \frac{d}{d u} (\frac{u ^{3} - 1}{u ^{3} + 1})

= 8 (\frac{u ^{3} - 1}{u ^{3} + 1})^{7} \frac{d}{d u} (\frac{u ^{3} - 1}{u ^{3} + 1})

\frac{d}{d u} (\frac{u ^{3} - 1}{u ^{3} + 1}) = \frac{6 u ^{2}}{( u ^{3} + 1 ) ^{2}}

= 8 (\frac{u ^{3} - 1}{u ^{3} + 1})^{7} \frac{6 u ^{2}}{( u ^{3} + 1 ) ^{2}}

Vereinfache

8 (\frac{u ^{3} - 1}{u ^{3} + 1})^{7} \frac{6 u ^{2}}{( u ^{3} + 1 ) ^{2}} : \frac{48 u ^{2} ( u ^{3} - 1 ) ^{7}}{( u ^{3} + 1 ) ^{9}}

= \frac{48 u ^{2} ( u ^{3} - 1 ) ^{7}}{( u ^{3} + 1 ) ^{9}}

\frac{\partial}{\partial v} (uv)

18 y^{^{'}} = y (y^{2} - 9)

x \to 1 lim (\frac{1}{( 1.5 - 1 ) ^{2}})

\frac{\partial}{\partial x} (tan (4 x y^{5}))

\frac{d}{d x} (\frac{2 x ^{2} + x - 1}{x ^{2} - 1})