What is (\partial)/(\partial x)(ce^{-2x}+e^{-x}) ?

The answer to (\partial)/(\partial x)(ce^{-2x}+e^{-x}) is -2ce^{-2x}-e^{-x}

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(\partial)/(\partial x)(ce^{-2x}+e^{-x})

\frac{\partial}{\partial x} (c e^{- 2 x} + e^{- x})

- 2 c e^{- 2 x} - e^{- x}

Solution steps

\frac{\partial}{\partial x} (c e^{- 2 x} + e^{- x})

Apply the Sum/Difference Rule:

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

= \frac{\partial}{\partial x} (c e^{- 2 x}) + \frac{\partial}{\partial x} (e^{- x})

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

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

= - 2 c e^{- 2 x} - e^{- x}

\frac{d}{d x} (2 x^{4} + 4 x^{3} + x)

f (x) = \frac{x ^{2}}{1 + 2 x}

x \to 0 lim (1 - \frac{x ^{2}}{6})

\frac{t d u}{d t} = t^{2} + 3 u, t > 0, H (5) = 100

\frac{d}{d x} (x - 2 e^{x})

What is (\partial)/(\partial x)(ce^{-2x}+e^{-x}) ?
The answer to (\partial)/(\partial x)(ce^{-2x}+e^{-x}) is -2ce^{-2x}-e^{-x}