If you differentiate the equation of a curve, you will get a formula for the gradient of the curve.
Find the equation of the tangent to the curve y = x^3 at the point (2, 8).
dy/dx = 3x^2
Gradient of tangent when x = 2 is
3 × 2^2 = 12.
From the coordinate geometry section, the equation of the tangent is therefore:
y - 8 = 12(x - 2) since the gradient of the tangent is 12
and we know that it passes through (2, 8)
so y = 12x - 16