The slope of the tangent to the curve $y = \int_0^x \frac{1}{1+t^3} dt$ at the point where $x = 1$ is

The equation of the tangent to the curve $x y^5+2 x^2 y-x^3+y+1=0$ at $x=0$ is

If the tangent at point $(1, 2)$ on the curve $y = ax^2 + bx + \frac{7}{2}$ is parallel to the normal at $(-2, 2)$ on the curve $y = x^2 + 6x + 10$,then:

At which point is the tangent to the curve $y = x^3 + 5$ perpendicular to the line $x + 3y = 2$?

The point on the curve $y = \sqrt{x - 1}$ where the tangent is perpendicular to the line $2x + y - 5 = 0$ is

Show that the equation of the normal at any point on the curve $x=3 \cos \theta-\cos ^{3} \theta, y=3 \sin \theta-\sin ^{3} \theta$ is $4(y \cos ^{3} \theta-x \sin ^{3} \theta)=3 \sin 4 \theta$.