What is the derivative of ae^x+be^{-9x} ?

The derivative of ae^x+be^{-9x} is ae^x-9be^{-9x}

What is the first derivative of ae^x+be^{-9x} ?

The first derivative of ae^x+be^{-9x} is ae^x-9be^{-9x}

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\frac{d}{d x} (a e^{x} + b e^{- 9 x})

a e^{x} - 9 b e^{- 9 x}

Solution steps

\frac{d}{d x} (a e^{x} + b e^{- 9 x})

Apply the Sum/Difference Rule:

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

= \frac{d}{d x} (a e^{x}) + \frac{d}{d x} (b e^{- 9 x})

\frac{d}{d x} (a e^{x}) = a e^{x}

\frac{d}{d x} (b e^{- 9 x}) = - 9 b e^{- 9 x}

= a e^{x} - 9 b e^{- 9 x}

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What is the derivative of ae^x+be^{-9x} ?
The derivative of ae^x+be^{-9x} is ae^x-9be^{-9x}
What is the first derivative of ae^x+be^{-9x} ?
The first derivative of ae^x+be^{-9x} is ae^x-9be^{-9x}