Slope of the Tangent to Curve at Specified Point

To find the slope of the tangent to the curve $(y - x^5)^2 = x(1 + x^2)^2$ at the point $(1, 3)$, we need to determine the value of $\frac{dy}{dx}$ at $(x, y) = (1, 3)$.

First, verify that the point $(1, 3)$ lies on the curve: $\text{LHS} = (3 - 1^5)^2 = (2)^2 = 4$ $\text{RHS} = 1 \cdot (1 + 1^2)^2 = 1 \cdot (2)^2 = 4$ Since $\text{LHS} = \text{RHS}$, the point $(1, 3)$ lies on the curve.

Now, differentiate both sides of the given implicit equation with respect to $x$: $\frac{d}{dx} \left[ (y - x^5)^2 \right] = \frac{d}{dx} \left[ x(1 + x^2)^2 \right]$

Applying the chain rule to the left-hand side and the product rule along with the chain rule to the right-hand side gives: $2(y - x^5) \cdot \left( \frac{dy}{dx} - 5x^4 \right) = (1 + x^2)^2 \cdot \frac{d}{dx}(x) + x \cdot \frac{d}{dx} \left[ (1 + x^2)^2 \right]$

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

Simplifying the right-hand side: $2(y - x^5)\left( \frac{dy}{dx} - 5x^4 \right) = (1 + x^2)^2 + 4x^2(1 + x^2)$

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

Thus, the differentiated equation is: $2(y - x^5)\left( \frac{dy}{dx} - 5x^4 \right) = (1 + x^2)(1 + 5x^2)$

Substitute $x = 1$ and $y = 3$ into this equation: $2(3 - 1^5) \left( \left. \frac{dy}{dx} \right|_{(1,3)} - 5(1)^4 \right) = (1 + 1^2)(1 + 5(1)^2)$

$2(2) \left( \left. \frac{dy}{dx} \right|_{(1,3)} - 5 \right) = (2)(6)$

$4 \left( \left. \frac{dy}{dx} \right|_{(1,3)} - 5 \right) = 12$

Divide both sides by $4$: $\left. \frac{dy}{dx} \right|_{(1,3)} - 5 = 3$

$\left. \frac{dy}{dx} \right|_{(1,3)} = 8$

The slope of the tangent to the curve at the point $(1, 3)$ is $8$.