Determine $y^{\text{'}}(x)$ if $y(x)=(x^4{x}^3{)}^{\frac{1}{x}}.$

Select the correct answer below:

$y^{\text{'}}(x)=\frac{-1}{x^2}(ln(x^4{x}^3))(x^4{x}^3{)}^{\frac{1}{x}}$

$y^{\text{'}}(x)=(x^4{x}^3{)}^{\frac{1}{x}}(\frac{-1}{x^2}ln(x^4{x}^3)\frac{(4x^{3}3x^2)\frac{1}{x}}{x^4{x}^3})$

$y^{\text{'}}(x)=(x^4{x}^3{)}^{\frac{1}{x}}(\frac{-1}{x^2}ln(x^4{x}^3)\frac{1}{x(x^4{x}^3)})$

$y^{\text{'}}(x)=(\frac{1}{x})(x^4{x}^3{)}^{\frac{1}{x}-1}$