What is the d/(dy)((1/(y^2)-5/(y^4))(y+9y^3)) ?

The d/(dy)((1/(y^2)-5/(y^4))(y+9y^3)) is (44)/(y^2)+(15)/(y^4)+9

What is the first d/(dy)((1/(y^2)-5/(y^4))(y+9y^3)) ?

The first d/(dy)((1/(y^2)-5/(y^4))(y+9y^3)) is (44)/(y^2)+(15)/(y^4)+9

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\frac{d}{d y} ((\frac{1}{y ^{2}} - \frac{5}{y ^{4}}) (y + 9 y^{3}))

\frac{44}{y ^{2}} + \frac{15}{y ^{4}} + 9

Solution steps

\frac{d}{d y} ((\frac{1}{y ^{2}} - \frac{5}{y ^{4}}) (y + 9 y^{3}))

Apply the Product Rule:

(f \cdot g)^{'} = f^{'} \cdot g + f \cdot g^{'}

f = \frac{1}{y ^{2}} - \frac{5}{y ^{4}}, g = y + 9 y^{3}

= \frac{d}{d y} (\frac{1}{y ^{2}} - \frac{5}{y ^{4}}) (y + 9 y^{3}) + \frac{d}{d y} (y + 9 y^{3}) (\frac{1}{y ^{2}} - \frac{5}{y ^{4}})

\frac{d}{d y} (\frac{1}{y ^{2}} - \frac{5}{y ^{4}}) = - \frac{2}{y ^{3}} + \frac{20}{y ^{5}}

\frac{d}{d y} (y + 9 y^{3}) = 1 + 27 y^{2}

= (- \frac{2}{y ^{3}} + \frac{20}{y ^{5}}) (y + 9 y^{3}) + (1 + 27 y^{2}) (\frac{1}{y ^{2}} - \frac{5}{y ^{4}})

Simplify

(- \frac{2}{y ^{3}} + \frac{20}{y ^{5}}) (y + 9 y^{3}) + (1 + 27 y^{2}) (\frac{1}{y ^{2}} - \frac{5}{y ^{4}}) : \frac{44}{y ^{2}} + \frac{15}{y ^{4}} + 9

= \frac{44}{y ^{2}} + \frac{15}{y ^{4}} + 9

\frac{d y}{d x} = 2 - 3 y

\int_{0}^{1} \frac{2}{( x + 2 ) ^{2}} d x

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

\int \frac{- 2}{5 x ^{2}} d x

\frac{d y}{d x} - x y^{2} = 2 y x

What is the d/(dy)((1/(y^2)-5/(y^4))(y+9y^3)) ?
The d/(dy)((1/(y^2)-5/(y^4))(y+9y^3)) is (44)/(y^2)+(15)/(y^4)+9
What is the first d/(dy)((1/(y^2)-5/(y^4))(y+9y^3)) ?
The first d/(dy)((1/(y^2)-5/(y^4))(y+9y^3)) is (44)/(y^2)+(15)/(y^4)+9