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

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

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

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

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

Solution steps

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

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} (x)

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

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

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

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

Simplify

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

f (x) = 3 x^{2} - 4 x + 6, at x = 2

f (x) = x^{3} - 3 x + 4, (3, 22)

\frac{d}{d x} (1 + x^{4})

f (x) = \frac{x ^{2}}{x - 8}

\int 2 x^{2} y e^{x yz} d x

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