Find the exact gradient of the curve y=ln(1-cos2x) at the point with x-coordinate π/6

This is a past paper question for an A level OCR MEI paper for Maths.

We need to find the gradient of the curve so we know right away that we need to use differentiation.

The equation y = ln(1-cos2x) is difficult to differentiate by itself so we use the chain rule and a substitution. We say y = ln(u) where u = 1-cos2x.

dy/du = 1/u. This is something you just have to learn, that the differential of y= lnx is 1/x.

If you're stuck with finding du/dx, you can use another substitution if you want where u = 1-cosv and v = 2x, but most people know (after a lot of practice) that the differential of y=coskx is -ksinkx. I believe it's also in most formula books.

So here du/dx = -2sin2x (remember the change of sign when going from cos to sin when differentiating).

So from the chain rule, we know that dy/dx = dy/du x du/dx so dy/dx = 1/u x -2sin2x = -2sin2x/u and u = 1-cos2x so dy/dx = -2sin2x/(1-cos2x)

Now we need to find the value of this gradient when x = π/6

dy/dx = -2sin2x/(1-cos2x) = -2sin(2 x π/6)/(1-cos(2 x π/6)) = -2sin(π/3)/(1-cos(π/3)) 

Remember that we're working in radians here so for this, you need to put your calculator in radians too

We know that cos(π/3) = 1/2 (or 0.5, but 1/2 is easier for the moment) and sin(π/3) = root(3)/2

So dy/dx at x=π/3 gives (-2 x root(3)/2)/(1-1/2) => multiply top and bottom by 2 => -2root(3)/(2-1) = -2root(3)/1 = -2root(3)

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