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

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

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

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

y - \frac{1}{x + z}

Solution steps

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

Treat

y, x

as constants

Apply the Sum/Difference Rule:

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

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

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

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

= y - \frac{1}{x + z}

x \to - \infty lim (7)

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

\frac{d ^{4}}{d x ^{4}} (e^{\frac{2}{x}})

\int 3 x e^{7 x} d x

(y - (x^{2} + y^{2})) d x - x d y = 0

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