What is (\partial)/(\partial x)(1/(xyz)) ?

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

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(\partial)/(\partial x)(1/(xyz))

\frac{\partial}{\partial x} (\frac{1}{x yz})

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

Solution steps

\frac{\partial}{\partial x} (\frac{1}{x yz})

Treat

y, z

as constants

Take the constant out:

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

= \frac{1}{yz} \frac{\partial}{\partial x} (\frac{1}{x})

Apply exponent rule:

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

= \frac{1}{yz} \frac{\partial}{\partial x} (x^{- 1})

Apply the Power Rule:

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

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

Simplify

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

= - \frac{1}{x ^{2} yz}

f (x) = - 8 x^{2} + 7 x

(x - 8 y) d x + 4 x d y = 0

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

\frac{d}{d x} (x^{2} + c)

\frac{d}{d x} (\frac{ln ( x )}{x ^{7}})

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