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 -2/(y^3)-(27(-y^2-1))/(y^4)

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 -2/(y^3)-(27(-y^2-1))/(y^4)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

derivative of

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

- \frac{2}{y ^{3}} - \frac{27 ( - y ^{2} - 1 )}{y ^{4}}

Solution steps

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

Simplify

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

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

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

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

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

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 -2/(y^3)-(27(-y^2-1))/(y^4)
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 -2/(y^3)-(27(-y^2-1))/(y^4)