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

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

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

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

- \frac{3}{x ^{2} y ^{4}}

Solution steps

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

Treat

x

as a constant

Take the constant out:

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

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

Apply the Power Rule:

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

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

Simplify

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

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

\frac{1}{2} x 25 - x^{2}

\int (\frac{3 x ^{3} - 4 x - 2}{x ^{3}}) d x

f (x) = \frac{6 x}{x ^{2} + 1}, (1, 3)

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

\frac{d}{d x} (\frac{e ^{2 x}}{x ^{2} + 1})

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