What is the d/(du)(((2u^3+7)(3u^2-5))/(u^2+1)) ?

The d/(du)(((2u^3+7)(3u^2-5))/(u^2+1)) is (18u^6+20u^4-30u^2+112u)/((u^2+1)^2)

What is the first d/(du)(((2u^3+7)(3u^2-5))/(u^2+1)) ?

The first d/(du)(((2u^3+7)(3u^2-5))/(u^2+1)) is (18u^6+20u^4-30u^2+112u)/((u^2+1)^2)

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\frac{d}{d u} (\frac{( 2 u ^{3} + 7 ) ( 3 u ^{2} - 5 )}{u ^{2} + 1})

\frac{18 u ^{6} + 20 u ^{4} - 30 u ^{2} + 112 u}{( u ^{2} + 1 ) ^{2}}

Solution steps

\frac{d}{d u} (\frac{( 2 u ^{3} + 7 ) ( 3 u ^{2} - 5 )}{u ^{2} + 1})

Apply the Quotient Rule:

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

= \frac{\frac{d}{d u} ( ( 2 u ^{3} + 7 ) ( 3 u ^{2} - 5 ) ) ( u ^{2} + 1 ) - \frac{d}{d u} ( u ^{2} + 1 ) ( 2 u ^{3} + 7 ) ( 3 u ^{2} - 5 )}{( u ^{2} + 1 ) ^{2}}

\frac{d}{d u} ((2 u^{3} + 7) (3 u^{2} - 5)) = 30 u^{4} - 30 u^{2} + 42 u

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

= \frac{( 30 u ^{4} - 30 u ^{2} + 42 u ) ( u ^{2} + 1 ) - 2 u ( 2 u ^{3} + 7 ) ( 3 u ^{2} - 5 )}{( u ^{2} + 1 ) ^{2}}

Expand

(30 u^{4} - 30 u^{2} + 42 u) (u^{2} + 1) - 2 u (2 u^{3} + 7) (3 u^{2} - 5) : 18 u^{6} + 20 u^{4} - 30 u^{2} + 112 u

= \frac{18 u ^{6} + 20 u ^{4} - 30 u ^{2} + 112 u}{( u ^{2} + 1 ) ^{2}}

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

\frac{d}{d x} (- arctan (\frac{x}{y}))

\frac{d}{d x} (sin (x^{2} - x))

\int tan (x) sec^{6} (x) d x

y = (1 + 3 x)^{8}, (0, 1)

What is the d/(du)(((2u^3+7)(3u^2-5))/(u^2+1)) ?
The d/(du)(((2u^3+7)(3u^2-5))/(u^2+1)) is (18u^6+20u^4-30u^2+112u)/((u^2+1)^2)
What is the first d/(du)(((2u^3+7)(3u^2-5))/(u^2+1)) ?
The first d/(du)(((2u^3+7)(3u^2-5))/(u^2+1)) is (18u^6+20u^4-30u^2+112u)/((u^2+1)^2)