What is the derivative of 9e^{-x}+e^{4x} ?

The derivative of 9e^{-x}+e^{4x} is -9e^{-x}+e^{4x}*4

What is the first derivative of 9e^{-x}+e^{4x} ?

The first derivative of 9e^{-x}+e^{4x} is -9e^{-x}+e^{4x}*4

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\frac{d}{d x} (9 e^{- x} + e^{4 x})

- 9 e^{- x} + e^{4 x} \cdot 4

Solution steps

\frac{d}{d x} (9 e^{- x} + e^{4 x})

Apply the Sum/Difference Rule:

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

= \frac{d}{d x} (9 e^{- x}) + \frac{d}{d x} (e^{4 x})

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

\frac{d}{d x} (e^{4 x}) = e^{4 x} \cdot 4

= - 9 e^{- x} + e^{4 x} \cdot 4

\int (\frac{x ^{3}}{x ^{2} + 1}) d x

\int 2500 - x^{2} d x

\frac{d}{d x} ((2 x^{2} - 1)^{3} \cdot (1 - x^{2})^{- 3})

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

What is the derivative of 9e^{-x}+e^{4x} ?
The derivative of 9e^{-x}+e^{4x} is -9e^{-x}+e^{4x}*4
What is the first derivative of 9e^{-x}+e^{4x} ?
The first derivative of 9e^{-x}+e^{4x} is -9e^{-x}+e^{4x}*4