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

The (x^{2x})^' is 2x^{2x}(ln(x)+1)

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

The first (x^{2x})^' is 2x^{2x}(ln(x)+1)

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

2 x^{2 x} (ln (x) + 1)

Solution steps

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

Apply exponent rule:

a^{b} = e^{b l n (a)}

x^{2 x} = e^{2 x l n (x)}

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

Apply the chain rule:

e^{2 x l n (x)} (2 x ln (x))^{'}

= e^{2 x l n (x)} (2 x ln (x))^{'}

(2 x ln (x))^{'} = 2 (ln (x) + 1)

= e^{2 x l n (x)} \cdot 2 (ln (x) + 1)

e^{2 x l n (x)} = x^{2 x}

= 2 x^{2 x} (ln (x) + 1)

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