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

The first (2e^{2x})^' is 4e^{2x}

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(2e^{2x})^'

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

4 e^{2 x}

Solution steps

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

Take the constant out:

(a \cdot f)^{'} = a \cdot f^{'}

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

Apply the chain rule:

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

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

(2 x)^{'} = 2

= 2 e^{2 x} \cdot 2

Simplify

= 4 e^{2 x}

- 9 (6 x^{2} + 7)^{- 6}

(x^{2} + y^{2})^{2}

\frac{\partial}{\partial y} (3 z e^{x yz})

\frac{x ^{2}}{x + 8}

x^{^{'}} = e^{x}