What is (\partial)/(\partial x)(ln(2ye^{xy})) ?

The answer to (\partial)/(\partial x)(ln(2ye^{xy})) is y

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(\partial)/(\partial x)(ln(2ye^{xy}))

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

y

Solution steps

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

Simplify

ln (2 y e^{x y}) : ln (2 y) + y x

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

Treat

y

as a constant

Apply the Sum/Difference Rule:

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

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

\frac{\partial}{\partial x} (ln (2 y)) = 0

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

= 0 + y

Simplify

= y

\frac{d}{d x} (\frac{x + 1}{x - 1})

\frac{d}{d x} (x^{x^{x}})

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

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

\frac{3}{5 + x ^{2}}

What is (\partial)/(\partial x)(ln(2ye^{xy})) ?
The answer to (\partial)/(\partial x)(ln(2ye^{xy})) is y