What is (\partial)/(\partial z)(e^{xy}+ln(2xz)+sin(x^2y^2)) ?

The answer to (\partial)/(\partial z)(e^{xy}+ln(2xz)+sin(x^2y^2)) is 1/z

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(\partial)/(\partial z)(e^{xy}+ln(2xz)+sin(x^2y^2))

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

\frac{1}{z}

Solution steps

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

Treat

x, y

as constants

Apply the Sum/Difference Rule:

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

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

\frac{\partial}{\partial z} (e^{x y}) = 0

\frac{\partial}{\partial z} (ln (2 x z)) = \frac{1}{z}

\frac{\partial}{\partial z} (sin (x^{2} y^{2})) = 0

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

Simplify

= \frac{1}{z}

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

3 x^{2} + 3 x

\frac{d}{d x} (2 ln (x) - \frac{x ^{2}}{16})

\int \frac{x}{9 x ^{2} - 4} d x

y^{^{''}} + 6 y^{^{'}} + 34 y = 68, y (0) = 5, y^{^{'}} (0) = 7

What is (\partial)/(\partial z)(e^{xy}+ln(2xz)+sin(x^2y^2)) ?
The answer to (\partial)/(\partial z)(e^{xy}+ln(2xz)+sin(x^2y^2)) is 1/z