Answer:
Option A
Explanation:
Plan
Slope of tangent at the point $(x_{1},y_{1})$ is $ \left(\frac{dy}{dx}\right)_{(x_{1},y_{1})}$
Given curve $(y-x^{5})^{2}=x(1+x^{2})^{2}$
$\Rightarrow 2(y-x^{5})\left(\frac{\text{d}y}{\text{d}x}-5x^{4}\right)=(1+x^{2})^{2}+2x(1+x^{2}).2x$
Put x=1, y=3
$\therefore$ $\frac{dy}{dx}=8$
