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

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

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

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

e^{x y} x y^{2} + 2 y e^{x y} - \frac{x}{y ^{2}}

Solution steps

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

Treat

x

as a constant

Apply the Sum/Difference Rule:

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

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

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

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

= e^{x y} x y^{2} + 2 y e^{x y} - \frac{x}{y ^{2}}

\frac{\partial}{\partial r} (r^{2} + s^{2})

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

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

y = (7 - 3 x^{3})^{3}

\int 16 e^{x} d x

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