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

The answer to (\partial)/(\partial z)(yz^2-ln(x+z)) is 2yz-1/(x+z)

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

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

2 yz - \frac{1}{x + z}

Solution steps

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

Treat

y, x

as constants

Apply the Sum/Difference Rule:

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

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

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

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

= 2 yz - \frac{1}{x + z}

3 x^{6} - 6 x^{6} y^{7} + 4 x^{5} y^{4}

\int \frac{1}{( 4 x ^{2} - 1 ) ^{\frac{3}{2}}} d x

\int_{1}^{3} 5 x e^{x} d x

f (x) = 3 x^{6} (x^{4} - 7 x)

\int x^{2} - 25 d x

What is (\partial)/(\partial z)(yz^2-ln(x+z)) ?
The answer to (\partial)/(\partial z)(yz^2-ln(x+z)) is 2yz-1/(x+z)