What is (\partial)/(\partial x)(e^z+1/x+xe^{-y}) ?

The answer to (\partial)/(\partial x)(e^z+1/x+xe^{-y}) is -1/(x^2)+e^{-y}

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(\partial)/(\partial x)(e^z+1/x+xe^{-y})

\frac{\partial}{\partial x} (e^{z} + \frac{1}{x} + x e^{- y})

- \frac{1}{x ^{2}} + e^{- y}

Solution steps

\frac{\partial}{\partial x} (e^{z} + \frac{1}{x} + x e^{- y})

Treat

z, y

as constants

Apply the Sum/Difference Rule:

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

= \frac{\partial}{\partial x} (e^{z}) + \frac{\partial}{\partial x} (\frac{1}{x}) + \frac{\partial}{\partial x} (x e^{- y})

\frac{\partial}{\partial x} (e^{z}) = 0

\frac{\partial}{\partial x} (\frac{1}{x}) = - \frac{1}{x ^{2}}

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

= 0 - \frac{1}{x ^{2}} + e^{- y}

Simplify

= - \frac{1}{x ^{2}} + e^{- y}

\int 2^{5} d x

\frac{d y}{d x} = \frac{4 y ln ( x )}{x}, y (e) = 1

x \to - 2 lim (x^{3} + x - 1)

\frac{d}{d x} (\frac{7}{x + 1})

\frac{d}{d x} (x^{2} ln (\frac{x}{4}))

What is (\partial)/(\partial x)(e^z+1/x+xe^{-y}) ?
The answer to (\partial)/(\partial x)(e^z+1/x+xe^{-y}) is -1/(x^2)+e^{-y}