The slope of the tangent to the curve $(y - x^{5})^{2} = x(1 + x^{2})^{2}$ at the point $(1, 3)$ is

The slope of the tangent to the curve (y - x⁵)² = x(1 + x²)² at the point (1, 3) is

$(y-x^5)^2 = x(1+x^2)^2$

$\text{point } (1,3)$

$\text{slope} = \left.\dfrac{dy}{dx}\right|_{(1,3)}$

$1.\ \text{differentiate both sides in } x$

$2.\ \text{simplify the right side}$

$3.\ \text{put } (1,3),\ \text{solve for } y'$

$2(y-x^5)(y'-5x^4) = \dfrac{d}{dx}\left[x(1+x^2)^2\right]$

$= (1+x^2)^2 + x\cdot 2(1+x^2)(2x)$

$= (1+x^2)\left[(1+x^2)+4x^2\right] = (1+x^2)(1+5x^2)$

$\text{at } (1,3):\ y-x^5 = 2,\quad (1+x^2)(1+5x^2) = 2\cdot 6 = 12$

$2\cdot 2\,(y'-5) = 12$

$4(y'-5) = 12 \Rightarrow y'-5 = 3$

$y' = 8$

$\text{slope} = 8$

$\boxed{8}$