What is (\partial)/(\partial y)(xe^y-ln(x)) ?

The answer to (\partial)/(\partial y)(xe^y-ln(x)) is e^yx

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(\partial)/(\partial y)(xe^y-ln(x))

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

e^{y} x

Solution steps

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

Treat

x

as a constant

Apply the Sum/Difference Rule:

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

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

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

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

= x e^{y} - 0

Simplify

= e^{y} x

What is (\partial)/(\partial y)(xe^y-ln(x)) ?
The answer to (\partial)/(\partial y)(xe^y-ln(x)) is e^yx