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

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

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

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

- \frac{1}{z + x}

Solution steps

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

Treat

y, z

as constants

Apply the Sum/Difference Rule:

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

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

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

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

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

Simplify

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

f (x) = x^{2} 2 x + 4

x^{2} \frac{d y}{d x} + x y = 3

\frac{d}{d x} ((1 + cos (2 x))^{- 4})

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

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