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}

\frac{d}{d t} (\frac{d x}{d t}) + \frac{d}{d t} (6 x) + 10 x = 0

y + 4 y^{3} = (y^{4} + 2 x) y^{^{'}}

\int x^{2} cos (k x) d x

y^{^{'}} - 2 x y = x, y (0) = 0

\int 2^{2 x + 1} d x

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