What is (\partial)/(\partial z)(yz^2-ln(x+z)) ?

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

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(\partial)/(\partial z)(yz^2-ln(x+z))

\frac{\partial}{\partial z} (y z^{2} - ln (x + z))

2 yz - \frac{1}{x + z}

Solution steps

\frac{\partial}{\partial z} (y z^{2} - ln (x + z))

Treat

y, x

as constants

Apply the Sum/Difference Rule:

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

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

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

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

= 2 yz - \frac{1}{x + z}

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