What is the (e^{2x}+e^{-x})^' ?

The (e^{2x}+e^{-x})^' is e^{2x}*2-e^{-x}

What is the first (e^{2x}+e^{-x})^' ?

The first (e^{2x}+e^{-x})^' is e^{2x}*2-e^{-x}

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(e^{2 x} + e^{- x})^{^{'}}

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

Solution steps

(e^{2 x} + e^{- x})^{'}

Apply the Sum/Difference Rule:

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

= (e^{2 x})^{^{'}} + (e^{- x})^{^{'}}

(e^{2 x})^{'} = e^{2 x} \cdot 2

(e^{- x})^{'} = - e^{- x}

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

y = - 3 - 5 x^{2}

t cos (3 t)

\int - x^{\frac{2}{3}} d x

\int_{0}^{24} 6 x d x

\frac{d}{d x} (3 x ln (2 x))

What is the (e^{2x}+e^{-x})^' ?
The (e^{2x}+e^{-x})^' is e^{2x}*2-e^{-x}
What is the first (e^{2x}+e^{-x})^' ?
The first (e^{2x}+e^{-x})^' is e^{2x}*2-e^{-x}