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

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

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

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

z^{2}

Solution steps

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

Treat

z, x

as constants

Apply the Sum/Difference Rule:

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

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

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

\frac{\partial}{\partial y} (ln (x + z)) = 0

= z^{2} - 0

Simplify

= z^{2}

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\frac{d}{d x} (sinh (x^{2}))

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