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

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

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

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

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derivative of

y = 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

x \to 0 lim (1 - \frac{cos ( x )}{x})

\frac{5 + x - 5}{x}

d y - (y - 8)^{2} d x = 0

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

x y^{^{'}} + y = - 3 x y^{2}, y (1) = 4

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