What is (\partial)/(\partial y)(ye^{xy}-1/y) ?

The answer to (\partial)/(\partial y)(ye^{xy}-1/y) is e^{yx}+e^{yx}yx+1/(y^2)

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(\partial)/(\partial y)(ye^{xy}-1/y)

\frac{\partial}{\partial y} (y e^{x y} - \frac{1}{y})

e^{y x} + e^{y x} y x + \frac{1}{y ^{2}}

Solution steps

\frac{\partial}{\partial y} (y e^{x y} - \frac{1}{y})

Treat

x

as a constant

Apply the Sum/Difference Rule:

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

= \frac{\partial}{\partial y} (y e^{x y}) - \frac{\partial}{\partial y} (\frac{1}{y})

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

\frac{\partial}{\partial y} (\frac{1}{y}) = - \frac{1}{y ^{2}}

= e^{x y} + x e^{x y} y - (- \frac{1}{y ^{2}})

Simplify

= e^{y x} + e^{y x} y x + \frac{1}{y ^{2}}

\frac{d}{d x} (4 x^{3} - 66 x^{2} + 216 x)

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

\frac{x ^{3}}{6} - 3 x^{2} + \frac{( 77 x )}{6} - 8, at x = 5

t^{3} ln (t)

\frac{d y}{d x} = \frac{y}{2 + x ^{2}}

What is (\partial)/(\partial y)(ye^{xy}-1/y) ?
The answer to (\partial)/(\partial y)(ye^{xy}-1/y) is e^{yx}+e^{yx}yx+1/(y^2)