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

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

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

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

- e^{- x}

Solution steps

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

Treat

y

as a constant

Apply the Sum/Difference Rule:

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

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

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

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

= 0 - e^{- x}

Simplify

= - e^{- x}

f (x) = 4 sin (4 x)

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

x^{^{'}} + 6 x = 0

n = 0 \sum \infty 3 \cdot (- \frac{2}{5})^{n}

f (x) = \frac{2 x}{x - 1}, at x = 2

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