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

The d/(dy)((1/(y^2)-3/(y^4))(y+5y^3)) is (14)/(y^2)+9/(y^4)+5

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

The first d/(dy)((1/(y^2)-3/(y^4))(y+5y^3)) is (14)/(y^2)+9/(y^4)+5

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

\frac{14}{y ^{2}} + \frac{9}{y ^{4}} + 5

Solution steps

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

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

\frac{d}{d y} (\frac{1}{y ^{2}} - \frac{3}{y ^{4}}) = - \frac{2}{y ^{3}} + \frac{12}{y ^{5}}

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

= (- \frac{2}{y ^{3}} + \frac{12}{y ^{5}}) (y + 5 y^{3}) + (1 + 15 y^{2}) (\frac{1}{y ^{2}} - \frac{3}{y ^{4}})

Simplify

(- \frac{2}{y ^{3}} + \frac{12}{y ^{5}}) (y + 5 y^{3}) + (1 + 15 y^{2}) (\frac{1}{y ^{2}} - \frac{3}{y ^{4}}) : \frac{14}{y ^{2}} + \frac{9}{y ^{4}} + 5

= \frac{14}{y ^{2}} + \frac{9}{y ^{4}} + 5

y^{^{''}} - y^{^{'}} - 2 y = 5 e^{- t}

\frac{d}{d x} (1 - x ln (x))

e^{2} t^{2}

sin (x) + x cos (x)

What is the d/(dy)((1/(y^2)-3/(y^4))(y+5y^3)) ?
The d/(dy)((1/(y^2)-3/(y^4))(y+5y^3)) is (14)/(y^2)+9/(y^4)+5
What is the first d/(dy)((1/(y^2)-3/(y^4))(y+5y^3)) ?
The first d/(dy)((1/(y^2)-3/(y^4))(y+5y^3)) is (14)/(y^2)+9/(y^4)+5