What is (\partial)/(\partial y)(4x^3y^{-2}) ?

The answer to (\partial)/(\partial y)(4x^3y^{-2}) is -(8x^3)/(y^3)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

(\partial)/(\partial y)(4x^3y^{-2})

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

- \frac{8 x ^{3}}{y ^{3}}

Solution steps

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

Treat

x

as a constant

Take the constant out:

(a \cdot f)^{'} = a \cdot f^{'}

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

Apply the Power Rule:

\frac{d}{d x} (x^{a}) = a \cdot x^{a - 1}

= 4 x^{3} (- 2 y^{- 2 - 1})

Simplify

4 x^{3} (- 2 y^{- 2 - 1}) : - \frac{8 x ^{3}}{y ^{3}}

= - \frac{8 x ^{3}}{y ^{3}}

t e^{- t} cos (t)

\int 0. 8^{x} d x

\int (3 x^{2} - 2 x + 1) (x^{3} - x^{2} + x)^{9} d x

\int_{0}^{5} x + 1 d x

What is (\partial)/(\partial y)(4x^3y^{-2}) ?
The answer to (\partial)/(\partial y)(4x^3y^{-2}) is -(8x^3)/(y^3)