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

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

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

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

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

Solution steps

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

Treat

x

as a constant

Apply the Sum/Difference Rule:

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

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

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

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

= 0 + e^{x y} + x e^{x y} y

Simplify

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

t \to 0 lim (\frac{6}{t} - \frac{6}{t ^{2} + t})

\int - 2^{- x} d x

n = 1 \sum \infty \frac{1}{3 ^{n - 1}}

\frac{d}{d x} (\frac{16}{x})

y^{^{''}} - 3 y^{^{'}} = - 9

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