ANSWER
[tex]\frac{dy}{dx} = x{(3x)}^{ {x}^{2} } (ln(9 {x}^{2} ) + 1)[/tex]
EXPLANATION
The given equation is;
[tex]y = {(3x)}^{ {x}^{2} } [/tex]
Take natural log of both sides:
[tex] ln(y) = ln({(3x)}^{ {x}^{2} } )[/tex]
This implies that,
[tex]ln(y) = {x}^{2} ln({(3x)})[/tex]
We differentiate to get,
[tex] \frac{1}{y} \frac{dy}{dx} = ln(3x) (2x) + {x}^{2} ( \frac{1}{x} )[/tex]
This simplifies to
[tex] \frac{dy}{dx} =y (2x ln(3x) + x)[/tex]We substitute for y to get:
[tex]\frac{dy}{dx} = {(3x)}^{ {x}^{2} } (2x ln(3x) + x)[/tex]
Or
[tex]\frac{dy}{dx} = x{(3x)}^{ {x}^{2} } (2ln(3x) + 1)[/tex]
[tex]\frac{dy}{dx} = x{(3x)}^{ {x}^{2} } (ln(9 {x}^{2} ) + 1)[/tex]
