What is the derivative of g(u)=(8u^2)/((u^2+u)^3) ?

The derivative of g(u)=(8u^2)/((u^2+u)^3) is (8(-4u-1))/(u^2(u+1)^4)

What is the first derivative of g(u)=(8u^2)/((u^2+u)^3) ?

The first derivative of g(u)=(8u^2)/((u^2+u)^3) is (8(-4u-1))/(u^2(u+1)^4)

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derivative of

g (u) = \frac{8 u ^{2}}{( u ^{2} + u ) ^{3}}

\frac{8 ( - 4 u - 1 )}{u ^{2} ( u + 1 ) ^{4}}

Solution steps

\frac{d}{d u} (\frac{8 u ^{2}}{( u ^{2} + u ) ^{3}})

Take the constant out:

(a \cdot f)^{'} = a \cdot f^{'}

= 8 \frac{d}{d u} (\frac{u ^{2}}{( u ^{2} + u ) ^{3}})

Apply the Quotient Rule:

(\frac{f}{g})^{'} = \frac{f ^{'} \cdot g - g ^{'} \cdot f}{g ^{2}}

= 8 \cdot \frac{\frac{d}{d u} ( u ^{2} ) ( u ^{2} + u ) ^{3} - \frac{d}{d u} ( ( u ^{2} + u ) ^{3} ) u ^{2}}{( ( u ^{2} + u ) ^{3} ) ^{2}}

\frac{d}{d u} (u^{2}) = 2 u

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

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

Simplify

8 \cdot \frac{2 u ( u ^{2} + u ) ^{3} - 3 ( u ^{2} + u ) ^{2} ( 2 u + 1 ) u ^{2}}{( ( u ^{2} + u ) ^{3} ) ^{2}} : \frac{8 ( - 4 u - 1 )}{u ^{2} ( u + 1 ) ^{4}}

= \frac{8 ( - 4 u - 1 )}{u ^{2} ( u + 1 ) ^{4}}

\int - x^{2} - 2 x + 3 d x

\int \frac{2 x + 1}{x ^{2} + x + 3} d x

f (x) = \frac{x ^{2} - 81}{x - 9}

\int sin (12 x) cos (3 x) d x

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

What is the derivative of g(u)=(8u^2)/((u^2+u)^3) ?
The derivative of g(u)=(8u^2)/((u^2+u)^3) is (8(-4u-1))/(u^2(u+1)^4)
What is the first derivative of g(u)=(8u^2)/((u^2+u)^3) ?
The first derivative of g(u)=(8u^2)/((u^2+u)^3) is (8(-4u-1))/(u^2(u+1)^4)