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

The derivative of 1/(y^2)-5/(y^4)(y+9y^3) is -2/(y^3)-(15(-3y^2-1))/(y^4)

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

The first derivative of 1/(y^2)-5/(y^4)(y+9y^3) is -2/(y^3)-(15(-3y^2-1))/(y^4)

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derivative of

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

- \frac{2}{y ^{3}} - \frac{15 ( - 3 y ^{2} - 1 )}{y ^{4}}

Solution steps

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

Simplify

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

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

Apply the Sum/Difference Rule:

(f \pm g)^{'} = f^{'} \pm g^{'}

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

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

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

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

\frac{\partial}{\partial x} (sin (x y^{3}))

\int x^{3} (5 + x^{4})^{5} d x

8 (t + 1) \frac{d y}{d t} - 6 y = 12 t

\int_{0}^{1} (1 + 2^{x})^{2} 2^{- x} d x

\frac{\partial}{\partial x} (\cdot e)

What is the derivative of 1/(y^2)-5/(y^4)(y+9y^3) ?
The derivative of 1/(y^2)-5/(y^4)(y+9y^3) is -2/(y^3)-(15(-3y^2-1))/(y^4)
What is the first derivative of 1/(y^2)-5/(y^4)(y+9y^3) ?
The first derivative of 1/(y^2)-5/(y^4)(y+9y^3) is -2/(y^3)-(15(-3y^2-1))/(y^4)