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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

(\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}

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