What is the d/(dy)((1/(y^2)-7/(y^4))(y+5y^3)) ?

The d/(dy)((1/(y^2)-7/(y^4))(y+5y^3)) is (34)/(y^2)+(21)/(y^4)+5

What is the first d/(dy)((1/(y^2)-7/(y^4))(y+5y^3)) ?

The first d/(dy)((1/(y^2)-7/(y^4))(y+5y^3)) is (34)/(y^2)+(21)/(y^4)+5

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\frac{d}{d y} ((\frac{1}{y ^{2}} - \frac{7}{y ^{4}}) (y + 5 y^{3}))

\frac{34}{y ^{2}} + \frac{21}{y ^{4}} + 5

Solution steps

\frac{d}{d y} ((\frac{1}{y ^{2}} - \frac{7}{y ^{4}}) (y + 5 y^{3}))

Apply the Product Rule:

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

f = \frac{1}{y ^{2}} - \frac{7}{y ^{4}}, g = y + 5 y^{3}

= \frac{d}{d y} (\frac{1}{y ^{2}} - \frac{7}{y ^{4}}) (y + 5 y^{3}) + \frac{d}{d y} (y + 5 y^{3}) (\frac{1}{y ^{2}} - \frac{7}{y ^{4}})

\frac{d}{d y} (\frac{1}{y ^{2}} - \frac{7}{y ^{4}}) = - \frac{2}{y ^{3}} + \frac{28}{y ^{5}}

\frac{d}{d y} (y + 5 y^{3}) = 1 + 15 y^{2}

= (- \frac{2}{y ^{3}} + \frac{28}{y ^{5}}) (y + 5 y^{3}) + (1 + 15 y^{2}) (\frac{1}{y ^{2}} - \frac{7}{y ^{4}})

Simplify

(- \frac{2}{y ^{3}} + \frac{28}{y ^{5}}) (y + 5 y^{3}) + (1 + 15 y^{2}) (\frac{1}{y ^{2}} - \frac{7}{y ^{4}}) : \frac{34}{y ^{2}} + \frac{21}{y ^{4}} + 5

= \frac{34}{y ^{2}} + \frac{21}{y ^{4}} + 5

\frac{1}{( s ^{2} - 2 s )}

\int_{0}^{5} \frac{1}{x ^{2} - 6 x + 5} d x

x \to 3 + lim (ln x^{2} - 9)

\frac{d ^{3}}{d x ^{3}} (6 x + 8)

(1 - e^{t} + 3 e^{- 4 t}) cos (5 t)

What is the d/(dy)((1/(y^2)-7/(y^4))(y+5y^3)) ?
The d/(dy)((1/(y^2)-7/(y^4))(y+5y^3)) is (34)/(y^2)+(21)/(y^4)+5
What is the first d/(dy)((1/(y^2)-7/(y^4))(y+5y^3)) ?
The first d/(dy)((1/(y^2)-7/(y^4))(y+5y^3)) is (34)/(y^2)+(21)/(y^4)+5