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

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

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

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

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

Solution steps

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

Treat

y

as a constant

Apply the Sum/Difference Rule:

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

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

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

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

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

Simplify

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

f (x) = (\frac{480 t}{( t ^{2} + 5 ) ^{3}})

\int \frac{x ( y ^{2} - 5 y + 6 )}{x ^{2} + 1} d x

x \to 5 lim (in (x^{2} - 25))

\int \frac{1}{x + 1 - x} d x

x \to 1 lim (\frac{( x )}{x - 1})

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