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

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

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

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

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derivative of

(\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 x} (4 - 2 xx)

in v erse l a pl a ce \frac{1}{s ^{2} + 2 s + 1}

\int sin^{4} (5 x) d x

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

\int \frac{x}{( x ^{2} + 4 ) ^{2}} d x

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