What is (\partial)/(\partial x)((z)(e^{xyz}-x-y)) ?

The answer to (\partial)/(\partial x)((z)(e^{xyz}-x-y)) is z(e^{zxy}zy-1)

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(\partial)/(\partial x)((z)(e^{xyz}-x-y))

\frac{\partial}{\partial x} ((z) (e^{x yz} - x - y))

z (e^{z x y} zy - 1)

Solution steps

\frac{\partial}{\partial x} ((z) (e^{x yz} - x - y))

Treat

z, y

as constants

Take the constant out:

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

= z \frac{\partial}{\partial x} (e^{x yz} - x - y)

Apply the Sum/Difference Rule:

(f \pm g)^{'} = f^{'} \pm g^{'}

= z (\frac{\partial}{\partial x} (e^{x yz}) - \frac{\partial}{\partial x} (x) - \frac{\partial}{\partial x} (y))

\frac{\partial}{\partial x} (e^{x yz}) = e^{x yz} zy

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

\frac{\partial}{\partial x} (y) = 0

= z (e^{x yz} zy - 1 - 0)

Simplify

= z (e^{z x y} zy - 1)

\frac{x ^{3} - x}{( x + 2 ) ( x - 2 )}

\frac{2 ( - 3 x ^{2} + 1 )}{( x ^{2} + 1 ) ^{3}}

- \frac{1}{8} (x + 2)^{2} (x - 4)^{2}

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

(2 - 2)^{5}

What is (\partial)/(\partial x)((z)(e^{xyz}-x-y)) ?
The answer to (\partial)/(\partial x)((z)(e^{xyz}-x-y)) is z(e^{zxy}zy-1)