What is (\partial)/(\partial z)(1/(x^2yz)) ?

The answer to (\partial)/(\partial z)(1/(x^2yz)) is -1/(x^2yz^2)

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(\partial)/(\partial z)(1/(x^2yz))

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

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

Solution steps

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

Treat

x, y

as constants

Take the constant out:

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

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

Apply exponent rule:

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

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

Apply the Power Rule:

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

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

Simplify

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

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

\int 9 + 4 x^{2} d x

\frac{( s + 2 )}{s ^{2} + s + 1} - \frac{1}{s}

5.2 x + 2.3

\int \frac{d y}{y} d x

(e^{c o s (x)})^{^{'}}

What is (\partial)/(\partial z)(1/(x^2yz)) ?
The answer to (\partial)/(\partial z)(1/(x^2yz)) is -1/(x^2yz^2)