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

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

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

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

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derivative of

(\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

\int_{2}^{4} f (x) d x

f^{^{'}} (x) = \frac{4}{x}

\frac{d}{d t} (sin (6 t))

\int \frac{1}{36 x ^{2} + 2} d x

\int_{0}^{\frac{1}{2}} cos^{3} (π x) d x

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