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

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

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

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

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

Solution steps

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

Treat

y

as a constant

Apply the Sum/Difference Rule:

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

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

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

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

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

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