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

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

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

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

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

Solution steps

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

Treat

x, y

as constants

Apply the Sum/Difference Rule:

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

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

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

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

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

64 y^{^{''}} + 80 y^{^{'}} + 25 y = 0

y^{^{''}} + 4 y = x

\int \frac{4 x ^{3} + 2}{x ^{2}} d x

\frac{d ^{2}}{d x ^{2}} (sin (5 x))

y = cot (1 - 2 x^{2})

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