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

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

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

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

\frac{1}{x + z}

Solution steps

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

Treat

z, y

as constants

Apply the Sum/Difference Rule:

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

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

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

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

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

Simplify

= \frac{1}{x + z}

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