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

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

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

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

e^{- x} (1 - e^{- y})

Solution steps

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

Treat

y

as a constant

Take the constant out:

(a \cdot f)^{'} = a \cdot f^{'}

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

Apply the Sum/Difference Rule:

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

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

\frac{\partial}{\partial x} (1) = 0

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

= (1 - e^{- y}) (0 - (- e^{- x}))

Simplify

= e^{- x} (1 - e^{- y})

- csc^{2} (x)

\frac{d y}{d x} = 9 x + y - 9

\int \frac{4 y ^{2} - 6 y + x + 2}{x}

y^{^{''}} - 2 y^{^{'}} = 0

\frac{\partial}{\partial x} ((x + y) ln (x + y))

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