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

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

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

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

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derivative of

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

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

Solution steps

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

Simplify

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

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

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

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

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

\int 3 x^{- 3} + sin (x) d x

\frac{d y}{d x} = \frac{y}{x} + \frac{x}{y}

\frac{\partial}{\partial θ} (r (θ) \cdot cos (θ))

e^{r t}

\int_{- π}^{π} sin^{3} (x) d x

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