use logarithmic differentiation to find the derivative of function

use logarithmic differentiation to find the derivative of function:

y=(x^2+2)^2(x^4+4)^4

$\small{ \begin{array}{|rcll|} \hline y &=& (x^2+2)^2(x^4+4)^4 \quad & | \quad \ln() \text{ both sides} \\ \ln(y) &=& \ln(~(x^2+2)^2(x^4+4)^4~) \\ \ln(y) &=& \ln(~(x^2+2)^2~) + \ln(~(x^4+4)^4~) \\ \ln(y) &=& 2\cdot\ln(x^2+2) + 4\cdot \ln(x^4+4) \quad & | \quad \text{differentiation both sides}\\ (~\ln(y)~)' &=& 2\cdot (~\ln(x^2+2)~)' + 4\cdot (~(\ln(x^4+4)~)' \\ \frac{y'}{y} &=& 2\cdot \frac{2x}{x^2+2} + 4\cdot \frac{4x^3}{x^4+4} \\ \frac{y'}{y} &=& \frac{4x}{x^2+2} + \frac{16x^3}{x^4+4} \\ y' &=& y\cdot \left( \frac{4x}{x^2+2} + \frac{16x^3}{x^4+4} \right) \quad & | \quad y = (x^2+2)^2(x^4+4)^4 \\ y' &=& (x^2+2)^2(x^4+4)^4 \cdot \left( \frac{4x}{x^2+2} + \frac{16x^3}{x^4+4} \right) \\ y' &=& 4x\cdot(x^2+2)^2(x^4+4)^4 \cdot \left( \frac{1}{x^2+2} + \frac{4x^2}{x^4+4} \right) \\ y' &=& 4x\cdot(x^2+2)(x^4+4)^3 \cdot \left[ \frac{ (x^2+2)(x^4+4)}{x^2+2} + \frac{4x^2\cdot(x^2+2)(x^4+4)}{x^4+4} \right] \\ y' &=& 4x\cdot(x^2+2)(x^4+4)^3 \cdot \left[ x^4+4 + 4x^2\cdot(x^2+2) \right] \\ y' &=& 4x\cdot(x^2+2)(x^4+4)^3 \cdot ( x^4+4 + 4x^4+ 8x^2 ) \\ \mathbf{y'} & \mathbf{=} & \mathbf{ 4x\cdot(x^2+2)(x^4+4)^3 \cdot ( 5x^4 +8x^2+4 ) }\\ \hline \end{array} }$