What is the (x^{3x})^' ?

The (x^{3x})^' is 3x^{3x}(ln(x)+1)

What is the first (x^{3x})^' ?

The first (x^{3x})^' is 3x^{3x}(ln(x)+1)

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(x^{3 x})^{^{'}}

3 x^{3 x} (ln (x) + 1)

Solution steps

(x^{3 x})^{'}

Apply exponent rule:

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

x^{3 x} = e^{3 x l n (x)}

= (e^{3 x l n (x)})^{^{'}}

Apply the chain rule:

e^{3 x l n (x)} (3 x ln (x))^{'}

= e^{3 x l n (x)} (3 x ln (x))^{'}

(3 x ln (x))^{'} = 3 (ln (x) + 1)

= e^{3 x l n (x)} \cdot 3 (ln (x) + 1)

e^{3 x l n (x)} = x^{3 x}

= 3 x^{3 x} (ln (x) + 1)