What is the derivative of y=9e^{-x}+e^{2x} ?

The derivative of y=9e^{-x}+e^{2x} is -9e^{-x}+e^{2x}*2

What is the first derivative of y=9e^{-x}+e^{2x} ?

The first derivative of y=9e^{-x}+e^{2x} is -9e^{-x}+e^{2x}*2

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derivative of

y = 9 e^{- x} + e^{2 x}

- 9 e^{- x} + e^{2 x} \cdot 2

Solution steps

\frac{d}{d x} (9 e^{- x} + e^{2 x})

Apply the Sum/Difference Rule:

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

= \frac{d}{d x} (9 e^{- x}) + \frac{d}{d x} (e^{2 x})

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

\frac{d}{d x} (e^{2 x}) = e^{2 x} \cdot 2

= - 9 e^{- x} + e^{2 x} \cdot 2

\int 8 x^{3} - cos (4 x) d x

e

x^{2} \frac{d y}{d x} = x y + y^{2}

f (x) = \frac{3}{x}, (- 8, - \frac{3}{8})

\frac{d}{d h} (\frac{e ^{ah} - 1}{h})

What is the derivative of y=9e^{-x}+e^{2x} ?
The derivative of y=9e^{-x}+e^{2x} is -9e^{-x}+e^{2x}*2
What is the first derivative of y=9e^{-x}+e^{2x} ?
The first derivative of y=9e^{-x}+e^{2x} is -9e^{-x}+e^{2x}*2