What is (\partial)/(\partial y)(yln(y)-e^{-xy}) ?

The answer to (\partial)/(\partial y)(yln(y)-e^{-xy}) is ln(y)+1+e^{-yx}x

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(\partial)/(\partial y)(yln(y)-e^{-xy})

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

ln (y) + 1 + e^{- y x} x

Solution steps

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

Treat

x

as a constant

Apply the Sum/Difference Rule:

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

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

\frac{\partial}{\partial y} (y ln (y)) = ln (y) + 1

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

= ln (y) + 1 - (- x e^{- x y})

Simplify

= ln (y) + 1 + e^{- y x} x

\int \frac{3 x + 1}{x ^{2} - 10 x + 28} d x

200 (e)^{0.03 x}

y = x^{4} - 3 x^{2} + 3, y = x^{2}

y = 5 x^{2} - x^{3}, (1, 4)

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