Differentiate the function with respect to $x$: $(x+3)^{2} \cdot(x+4)^{3} \cdot(x+5)^{4}$

Let $y = (x+3)^{2} \cdot(x+4)^{3} \cdot(x+5)^{4}$.
Taking the natural logarithm on both sides,we get:
$\log y = \log((x+3)^{2} \cdot(x+4)^{3} \cdot(x+5)^{4})$
$\log y = 2 \log(x+3) + 3 \log(x+4) + 4 \log(x+5)$
Differentiating both sides with respect to $x$:
$\frac{1}{y} \frac{dy}{dx} = 2 \cdot \frac{1}{x+3} + 3 \cdot \frac{1}{x+4} + 4 \cdot \frac{1}{x+5}$
$\frac{dy}{dx} = y \left[ \frac{2}{x+3} + \frac{3}{x+4} + \frac{4}{x+5} \right]$
Substituting $y$ back:
$\frac{dy}{dx} = (x+3)^{2}(x+4)^{3}(x+5)^{4} \left[ \frac{2(x+4)(x+5) + 3(x+3)(x+5) + 4(x+3)(x+4)}{(x+3)(x+4)(x+5)} \right]$
$\frac{dy}{dx} = (x+3)(x+4)^{2}(x+5)^{3} [2(x^{2}+9x+20) + 3(x^{2}+8x+15) + 4(x^{2}+7x+12)]$
$\frac{dy}{dx} = (x+3)(x+4)^{2}(x+5)^{3} [2x^{2}+18x+40 + 3x^{2}+24x+45 + 4x^{2}+28x+48]$
$\frac{dy}{dx} = (x+3)(x+4)^{2}(x+5)^{3} (9x^{2}+70x+133)$