Find the slope of the tangent to the curve $y = x^{3} - x$ at $x = 2$.

The angle between the curves $x^2=8y$ and $xy=8$ is

$p_1$ and $p_2$ are the perpendicular distances from the origin to the tangent and normal drawn at any point on the curve $x^{2/3} + y^{2/3} = a^{2/3}$ respectively. If $k_1 p_1^2 + k_2 p_2^2 = a^2$,then $k_1 + k_2 =$

The curve $y=ax^3+bx^2+cx+5$ touches the $X$-axis at $P(-2,0)$ and cuts the $Y$-axis at a point $Q$,where its gradient is $3$. Then:

If the tangent drawn at the point $(x_1, y_1)$,where $x_1, y_1 \in \mathbb{N}$,on the curve $y = x^4 - 2x^3 + x^2 + 5x$ passes through the origin,then $x_1 + y_1 =$

If $x^2 + y^2 = R^2$ $(R > 0)$,then $k = \frac{y''}{\sqrt{(1 + (y')^2)^3}}$. The value of $k$ in terms of $R$ alone is equal to: