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

The d/(dy)((1/(y^2)-9/(y^4))(y+3y^3)) is (26)/(y^2)+(27)/(y^4)+3

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

The first d/(dy)((1/(y^2)-9/(y^4))(y+3y^3)) is (26)/(y^2)+(27)/(y^4)+3

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

\frac{26}{y ^{2}} + \frac{27}{y ^{4}} + 3

Solution steps

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

Apply the Product Rule:

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

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

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

\frac{d}{d y} (\frac{1}{y ^{2}} - \frac{9}{y ^{4}}) = - \frac{2}{y ^{3}} + \frac{36}{y ^{5}}

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

= (- \frac{2}{y ^{3}} + \frac{36}{y ^{5}}) (y + 3 y^{3}) + (1 + 9 y^{2}) (\frac{1}{y ^{2}} - \frac{9}{y ^{4}})

Simplify

(- \frac{2}{y ^{3}} + \frac{36}{y ^{5}}) (y + 3 y^{3}) + (1 + 9 y^{2}) (\frac{1}{y ^{2}} - \frac{9}{y ^{4}}) : \frac{26}{y ^{2}} + \frac{27}{y ^{4}} + 3

= \frac{26}{y ^{2}} + \frac{27}{y ^{4}} + 3

sin (10 t)

x \to 0 lim (\frac{8}{x} - \frac{8}{∣ x ∣})

\frac{d}{d x} (360 x^{2} - 480 x + 120)

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

\int \frac{e 3 x + 1}{3 x + 1} d x

What is the d/(dy)((1/(y^2)-9/(y^4))(y+3y^3)) ?
The d/(dy)((1/(y^2)-9/(y^4))(y+3y^3)) is (26)/(y^2)+(27)/(y^4)+3
What is the first d/(dy)((1/(y^2)-9/(y^4))(y+3y^3)) ?
The first d/(dy)((1/(y^2)-9/(y^4))(y+3y^3)) is (26)/(y^2)+(27)/(y^4)+3