What is (\partial)/(\partial x)(ye^x+xe^{-y}) ?

The answer to (\partial)/(\partial x)(ye^x+xe^{-y}) is ye^x+e^{-y}

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(\partial)/(\partial x)(ye^x+xe^{-y})

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

y e^{x} + e^{- y}

Solution steps

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

Treat

y

as a constant

Apply the Sum/Difference Rule:

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

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

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

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

= y e^{x} + e^{- y}

\frac{1}{( 7 + x ) ^{2}}

y = \frac{4 x}{x ^{2} + 1}, (1, 2)

\int 2 + t d t

What is (\partial)/(\partial x)(ye^x+xe^{-y}) ?
The answer to (\partial)/(\partial x)(ye^x+xe^{-y}) is ye^x+e^{-y}