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

8

The equation of the curve is

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

Differentiating both sides with respect to x, we get

$2\left(y-x^5\right)\left(\frac{d y}{d x}-5 x^4\right)=\left(1+x^2\right)^2+4 x^2\left(1+x^2\right)$

Putting x = 1, y = 3 on both sides, we get

$4\left(\frac{d y}{d x}-5\right)=4+8 \Rightarrow \frac{d y}{d x}=8$