By Differentiating from first principles, find the gradient of the curve f(x) = x^2 at the point where x = 2

This qustion can be solved easily using the gradient formular, m = ∆y/∆x, and some simple algrebra.

The gradient at the point x = 2 is calculated by find the gradient of a tangent at x = 2. To find this we imagine we are drawing a line from the point at x = 2, to a point on the line very close to it, a distance of dx along, and [f(dx) - f(x)] up.

Using the gradient m = ∆y/∆x, and subbing in values for change in x and change in y we get:

dy/dx = [f(x + dx) - f(x)]/dx

This however is the gradient along the line; the gradient at the point x = 2 is found by finding the limit as dx tends to zero, or as the line becomes infintesimently small.

This gives:

dy/dx = lim x -> 0 [(x + dx)^2 - x^2)/dx

= lim x -> 0 [x^2 + 2xdx + dx^2/dx

= lim x -> 0 [2x + dx]

= 2x

it is importnt to now finish the question and fin the gradient at x = 2

dy/dx = 2 x 2 = 4