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

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

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

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

z

Solution steps

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

Treat

z, x

as constants

Apply the Sum/Difference Rule:

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

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

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

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

= z - 0

Simplify

= z

x \to 2 lim (3 x^{2} + 1)

y = 300 (1.056)^{x}

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

n = 1 \sum \infty \frac{5}{π ^{2}}

f (x) = \frac{3}{x}, (7, \frac{3}{7})

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