What is (\partial)/(\partial x)(2y-xe^{xy}) ?

The answer to (\partial)/(\partial x)(2y-xe^{xy}) is -e^{yx}-e^{yx}yx

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(\partial)/(\partial x)(2y-xe^{xy})

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

- e^{y x} - e^{y x} y x

Solution steps

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

Treat

y

as a constant

Apply the Sum/Difference Rule:

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

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

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

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

= 0 - (e^{y x} + y e^{y x} x)

Simplify

= - e^{y x} - e^{y x} y x

n = 0 \sum \infty \frac{1}{n ^{\frac{2}{3}}}

\frac{d}{d x} (e^{(2 x)} (x^{3} + 6 x - 1) ln (x))

\frac{d}{d x} (\frac{sin ^{2} ( x ) + 1}{cos ^{2} ( x ) + 1})

\frac{x}{8}

{\frac{( s - 1 )}{s + 1}}

What is (\partial)/(\partial x)(2y-xe^{xy}) ?
The answer to (\partial)/(\partial x)(2y-xe^{xy}) is -e^{yx}-e^{yx}yx