Find the equation of the tangent to the curve `y=sec x` at the point `(0,1)`

The equation of the tangent to the curve y=sec x at the point (0,1) is (a) y=0 (b) x=0 (c) y=1 (d) x=1

Find the equation of the tangent to the curve y=-5x^2+6x+7 at the point (1//2,\ 35//4) .

Find the equation of the tangent to the curve y=(x^3-1)(x-2) at the points where the curve cuts the x-axis.

Find the equation of the tangents to the curve y=(x-1)(x-2) at the points where the curve cuts the x-axis.

Show that the normal to the curve 5x^(5)-10x^(3)+x+2y+6=0 at P(0,-3) meets the curve again at two points.Find the equations of the tangents to the curve at these points.

Find the equation of the tangent to the curve y=(x^(3)-1)(x-2) at the points where the curve cuts the x-axis.

Equation of the tangent to the curve y=e^(x) at (0,1) is

Find the equation of the tangent of the curve y=3x^(2) at (1, 1).

Equation of the tangent to the curve y= log x at (1,0) is

Find the equation of tangent to the curve y= x^(3)-x , at the point at which slope of tangent is equal to zero.