What is the derivative of (1/(y^2)-9/(y^4))(y+5y^3) ?

The derivative of (1/(y^2)-9/(y^4))(y+5y^3) is (44)/(y^2)+(27)/(y^4)+5

What is the first derivative of (1/(y^2)-9/(y^4))(y+5y^3) ?

The first derivative of (1/(y^2)-9/(y^4))(y+5y^3) is (44)/(y^2)+(27)/(y^4)+5

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derivative of

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

\frac{44}{y ^{2}} + \frac{27}{y ^{4}} + 5

Solution steps

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

= \frac{d}{d y} (\frac{1}{y ^{2}} - \frac{9}{y ^{4}}) (y + 5 y^{3}) + \frac{d}{d y} (y + 5 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 + 5 y^{3}) = 1 + 15 y^{2}

= (- \frac{2}{y ^{3}} + \frac{36}{y ^{5}}) (y + 5 y^{3}) + (1 + 15 y^{2}) (\frac{1}{y ^{2}} - \frac{9}{y ^{4}})

Simplify

(- \frac{2}{y ^{3}} + \frac{36}{y ^{5}}) (y + 5 y^{3}) + (1 + 15 y^{2}) (\frac{1}{y ^{2}} - \frac{9}{y ^{4}}) : \frac{44}{y ^{2}} + \frac{27}{y ^{4}} + 5

= \frac{44}{y ^{2}} + \frac{27}{y ^{4}} + 5

f (x) = ln (\frac{( 1 + x )}{( 1 - x )})

8 (ln (y)) y^{^{'}} - x^{7} y = 0

\frac{d}{d t} (cos (5 t))

\int_{- 2}^{2} x^{3} - 4 x d x

\int \frac{( 1 - x ^{2} ) ^{\frac{1}{2}}}{x ^{4}} d x

What is the derivative of (1/(y^2)-9/(y^4))(y+5y^3) ?
The derivative of (1/(y^2)-9/(y^4))(y+5y^3) is (44)/(y^2)+(27)/(y^4)+5
What is the first derivative of (1/(y^2)-9/(y^4))(y+5y^3) ?
The first derivative of (1/(y^2)-9/(y^4))(y+5y^3) is (44)/(y^2)+(27)/(y^4)+5