What is the (e^{7e^x})^' ?

The (e^{7e^x})^' is 7e^{7e^x+x}

What is the first (e^{7e^x})^' ?

The first (e^{7e^x})^' is 7e^{7e^x+x}

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(e^{7 e^{x}})^{^{'}}

7 e^{7 e^{x} + x}

Solution steps

(e^{7 e^{x}})^{'}

Apply the chain rule:

e^{7 e^{x}} (7 e^{x})^{'}

= e^{7 e^{x}} (7 e^{x})^{'}

(7 e^{x})^{'} = 7 e^{x}

= e^{7 e^{x}} \cdot 7 e^{x}

Apply exponent rule:

a^{b} \cdot a^{c} = a^{b + c}

e^{7 e^{x}} e^{x} = e^{7 e^{x} + x}

= 7 e^{7 e^{x} + x}

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