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

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

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

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

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

\frac{26}{y ^{2}} + \frac{9}{y ^{4}} + 9

Solution steps

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

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

= (- \frac{2}{y ^{3}} + \frac{12}{y ^{5}}) (y + 9 y^{3}) + (1 + 27 y^{2}) (\frac{1}{y ^{2}} - \frac{3}{y ^{4}})

Simplify

(- \frac{2}{y ^{3}} + \frac{12}{y ^{5}}) (y + 9 y^{3}) + (1 + 27 y^{2}) (\frac{1}{y ^{2}} - \frac{3}{y ^{4}}) : \frac{26}{y ^{2}} + \frac{9}{y ^{4}} + 9

= \frac{26}{y ^{2}} + \frac{9}{y ^{4}} + 9

2 csc (1 - 3 x)

f (x) = (2 x + 3)^{5}, at x = - 1

\frac{d y}{d x} - 2 y = 0

y^{^{''}} + 4 y^{^{'}} + 3 y = 3 x + 7

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