What is (\partial)/(\partial x)(e^{xy}+xye^{xy}) ?

The answer to (\partial)/(\partial x)(e^{xy}+xye^{xy}) is 2e^{xy}y+e^{xy}xy^2

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(\partial)/(\partial x)(e^{xy}+xye^{xy})

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

2 e^{x y} y + e^{x y} x y^{2}

Solution steps

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

Treat

y

as a constant

Apply the Sum/Difference Rule:

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

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

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

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

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

Simplify

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

= 2 e^{x y} y + e^{x y} x y^{2}

\frac{d}{d x} (\frac{a x ^{3} + b x + c}{d x ^{3} - e})

\int_{0}^{1} \frac{6 + x}{1 + x ^{2}} d x

f (x) = \frac{x ^{3}}{x + 1}, at x = 1

\frac{d}{d x} (tan (π x - 1) + e^{c o s (x)})

y = (x - 1)^{3}

What is (\partial)/(\partial x)(e^{xy}+xye^{xy}) ?
The answer to (\partial)/(\partial x)(e^{xy}+xye^{xy}) is 2e^{xy}y+e^{xy}xy^2