What is (\partial)/(\partial x)(ln(7ye^{xy})) ?

The answer to (\partial)/(\partial x)(ln(7ye^{xy})) is y

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(\partial)/(\partial x)(ln(7ye^{xy}))

\frac{\partial}{\partial x} (ln (7 y e^{x y}))

y

Solution steps

\frac{\partial}{\partial x} (ln (7 y e^{x y}))

Simplify

ln (7 y e^{x y}) : ln (7 y) + y x

= \frac{\partial}{\partial x} (ln (7 y) + y x)

Treat

y

as a constant

Apply the Sum/Difference Rule:

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

= \frac{\partial}{\partial x} (ln (7 y)) + \frac{\partial}{\partial x} (y x)

\frac{\partial}{\partial x} (ln (7 y)) = 0

\frac{\partial}{\partial x} (y x) = y

= 0 + y

Simplify

= y

f (x) = \frac{4 x}{x + 2}, (6, 3)

\int (- \frac{3}{x} + 3 e^{x}) d x

(- 10) (- 13.3)

\frac{1}{s ^{2} + 220}

x \to + 4 lim (\frac{x ^{2} - 1}{x - 4})

What is (\partial)/(\partial x)(ln(7ye^{xy})) ?
The answer to (\partial)/(\partial x)(ln(7ye^{xy})) is y