What is (\partial)/(\partial x)(2xe^y-xcos(xy)+2y) ?

The answer to (\partial)/(\partial x)(2xe^y-xcos(xy)+2y) is 2e^y-cos(xy)+xysin(xy)

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(\partial)/(\partial x)(2xe^y-xcos(xy)+2y)

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

2 e^{y} - cos (x y) + x y sin (x y)

Solution steps

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

Treat

y

as a constant

Apply the Sum/Difference Rule:

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

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

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

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

\frac{\partial}{\partial x} (2 y) = 0

= 2 e^{y} - (cos (y x) - y x sin (y x)) + 0

Simplify

= 2 e^{y} - cos (x y) + x y sin (x y)

y^{^{''}} + 45 y^{^{'}} + 20 y = 0

\int x x^{4} - 1 d x

\frac{d}{d x} (\frac{8 x ^{3} - 2 x ^{2} + 3}{2 x})

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

\int \frac{5 x - 4}{8 x - 5 ^{2} + 9} d x

What is (\partial)/(\partial x)(2xe^y-xcos(xy)+2y) ?
The answer to (\partial)/(\partial x)(2xe^y-xcos(xy)+2y) is 2e^y-cos(xy)+xysin(xy)