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

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

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

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

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

e^{x} - e^{- x}

Solution steps

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

Apply the Sum/Difference Rule:

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

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

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

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

= e^{x} - e^{- x}

x (x + 1) y^{^{'}} + x y = 1, y (e) = 1

\int (- \frac{2 x y}{( x ^{2} + y ^{2} )}) d y

\frac{\partial}{\partial x} ((1 + x) sin (y))

\frac{d}{d x} (ln (a x a + x))

\int \frac{1}{( x ^{2} - 49 ) ^{\frac{3}{2}}} d x