What is the derivative of g(u)=((u^3-1)/(u^3+1))^8 ?

The derivative of g(u)=((u^3-1)/(u^3+1))^8 is (48u^2(u^3-1)^7)/((u^3+1)^9)

What is the first derivative of g(u)=((u^3-1)/(u^3+1))^8 ?

The first derivative of g(u)=((u^3-1)/(u^3+1))^8 is (48u^2(u^3-1)^7)/((u^3+1)^9)

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derivative of

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

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

Solution steps

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

Apply the chain rule:

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}}

Simplify

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})

What is the derivative of g(u)=((u^3-1)/(u^3+1))^8 ?
The derivative of g(u)=((u^3-1)/(u^3+1))^8 is (48u^2(u^3-1)^7)/((u^3+1)^9)
What is the first derivative of g(u)=((u^3-1)/(u^3+1))^8 ?
The first derivative of g(u)=((u^3-1)/(u^3+1))^8 is (48u^2(u^3-1)^7)/((u^3+1)^9)