What is (\partial)/(\partial y)(ycos(x)+2xe^y) ?

The answer to (\partial)/(\partial y)(ycos(x)+2xe^y) is cos(x)+2xe^y

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(\partial)/(\partial y)(ycos(x)+2xe^y)

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

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

Solution steps

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

Treat

x

as a constant

Apply the Sum/Difference Rule:

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

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

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

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

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

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

\frac{d}{d x} (6 x^{\frac{3}{2}} + 1)

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

f (x) = arcsin (2 x + 1)

\frac{d}{d x} (x^{2} - 2 x + 2 (x + 1))

What is (\partial)/(\partial y)(ycos(x)+2xe^y) ?
The answer to (\partial)/(\partial y)(ycos(x)+2xe^y) is cos(x)+2xe^y