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

The answer to (\partial)/(\partial y)(cos(x-y)-xe^y) is sin(x-y)-xe^y

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

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

sin (x - y) - x e^{y}

Solution steps

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

Treat

x

as a constant

Apply the Sum/Difference Rule:

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

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

\frac{\partial}{\partial y} (cos (x - y)) = sin (x - y)

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

= sin (x - y) - x e^{y}

x^{2} + x y + 2 y^{2} = 16, (- 2, - 2)

\int (x^{2} - 2 x) (x^{3} - 3 x^{2})^{2} d x

\frac{\partial}{\partial x} ((5 x + 8 y)^{11})

\frac{d ^{2}}{d x ^{2}} (\frac{x ^{2}}{5 + x})

\int_{3}^{6} (2 x - 10) d x

What is (\partial)/(\partial y)(cos(x-y)-xe^y) ?
The answer to (\partial)/(\partial y)(cos(x-y)-xe^y) is sin(x-y)-xe^y