What is the derivative of 8e^{-x}+e^{3x} ?

The derivative of 8e^{-x}+e^{3x} is -8e^{-x}+e^{3x}*3

What is the first derivative of 8e^{-x}+e^{3x} ?

The first derivative of 8e^{-x}+e^{3x} is -8e^{-x}+e^{3x}*3

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derivative of

8 e^{- x} + e^{3 x}

- 8 e^{- x} + e^{3 x} \cdot 3

Solution steps

\frac{d}{d x} (8 e^{- x} + e^{3 x})

Apply the Sum/Difference Rule:

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

= \frac{d}{d x} (8 e^{- x}) + \frac{d}{d x} (e^{3 x})

\frac{d}{d x} (8 e^{- x}) = - 8 e^{- x}

\frac{d}{d x} (e^{3 x}) = e^{3 x} \cdot 3

= - 8 e^{- x} + e^{3 x} \cdot 3

\int - 2 tan (t) d t

\int tan^{5} (θ) sec^{3} (θ) d θ

\frac{\partial}{\partial x} (x^{3} z + z^{3} x - 2 yz)

f (x) = x^{2} x^{3} + 1, at x = 1

x \to 1 lim (x^{2} + 3 x - 2)

What is the derivative of 8e^{-x}+e^{3x} ?
The derivative of 8e^{-x}+e^{3x} is -8e^{-x}+e^{3x}*3
What is the first derivative of 8e^{-x}+e^{3x} ?
The first derivative of 8e^{-x}+e^{3x} is -8e^{-x}+e^{3x}*3