What is (\partial)/(\partial x)((z)(yz-ln(x+z))) ?

The answer to (\partial)/(\partial x)((z)(yz-ln(x+z))) is -z/(z+x)

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(\partial)/(\partial x)((z)(yz-ln(x+z)))

\frac{\partial}{\partial x} ((z) (yz - ln (x + z)))

- \frac{z}{z + x}

Solution steps

\frac{\partial}{\partial x} ((z) (yz - ln (x + z)))

Treat

z, y

as constants

Take the constant out:

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

= z \frac{\partial}{\partial x} (yz - ln (x + z))

Apply the Sum/Difference Rule:

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

= z (\frac{\partial}{\partial x} (yz) - \frac{\partial}{\partial x} (ln (x + z)))

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

\frac{\partial}{\partial x} (ln (x + z)) = \frac{1}{x + z}

= z (0 - \frac{1}{x + z})

Simplify

z (0 - \frac{1}{x + z}) : - \frac{z}{z + x}

= - \frac{z}{z + x}

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What is (\partial)/(\partial x)((z)(yz-ln(x+z))) ?
The answer to (\partial)/(\partial x)((z)(yz-ln(x+z))) is -z/(z+x)