What is (\partial)/(\partial y)(1/(xyz^2)) ?

The answer to (\partial)/(\partial y)(1/(xyz^2)) is -1/(xy^2z^2)

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(\partial)/(\partial y)(1/(xyz^2))

\frac{\partial}{\partial y} (\frac{1}{x y z ^{2}})

- \frac{1}{x y ^{2} z ^{2}}

Solution steps

\frac{\partial}{\partial y} (\frac{1}{x y z ^{2}})

Treat

x, z

as constants

Take the constant out:

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

= \frac{1}{x z ^{2}} \frac{\partial}{\partial y} (\frac{1}{y})

Apply exponent rule:

\frac{1}{a} = a^{- 1}

= \frac{1}{x z ^{2}} \frac{\partial}{\partial y} (y^{- 1})

Apply the Power Rule:

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

= \frac{1}{x z ^{2}} (- 1 \cdot y^{- 1 - 1})

Simplify

\frac{1}{x z ^{2}} (- 1 \cdot y^{- 1 - 1}) : - \frac{1}{x y ^{2} z ^{2}}

= - \frac{1}{x y ^{2} z ^{2}}

\int \frac{3}{x} - \frac{x x}{4} d x

\frac{d}{d x} (\frac{( x + 1 ) ( 2 x - 5 )}{2 x + 1})

8 x

\int \frac{x}{ln ( x )} d x

\int \frac{1}{5 - x} d x

What is (\partial)/(\partial y)(1/(xyz^2)) ?
The answer to (\partial)/(\partial y)(1/(xyz^2)) is -1/(xy^2z^2)