Find the exact gradient of the curve y = ln(1-cos 2x) at the point with x-coordinate π/6.

The gradient to the curve is given by the derivative of the function y = f(x) = ln(1 - cos 2x). This function is a composition of two other functions so we need to use the chain rule to find the derivative. Suppose we call g(x) = ln x and h(x) = 1 - cos 2x, then it is clear that f(x) = g(h(x)). The chain rule states that f'(x) = g'(h(x))h'(x) so we need to differentiate both g(x) and h(x).We have that g'(x) = 1/x and h'(x) = 0 - (-sin 2x) (2) = 2 sin 2x [by another simpler application of the chain rule]. Therefore we find that f'(x) = g'(h(x))h'(x) = 1/(1 - cos 2x) * 2 sin 2x = (2 sin 2x) / (1 - cos 2x). To find the gradient when the curve intersects the line x = π/6, we simply need to input the value x = π/6 into our derivative function; that is f'(π/6) = (2 sin π/3 ) / (1 - cos π/3 ) = ( 2 (√3) /2 ) / (1 - 1/2 ) = (√3) / (1/2) = 2√3. [These values of sin π/3 and cos π/3 can be easily derived using an equilateral triangle].