To sketch a derivative graph, just work out the gradients at the major points on the original.

In particular: A turning point on f(x) will always be an x intercept on f'(x) (as the gradient at a turning point is 0.)

Whenever the graph of f(x) is going down (ie the gradient of f(x) is negative), f'(x) will be negative and vica versa.

Also, remember that the derivative will always be one polynomial less than the original, so the derivative graph of a quartic will look like a cubic, the derivative graph of a cubic will look like a parabola, etc.

Once you know all that information, you should be able to sketch derivative graphs simply by looking at the original graph. Like Colin said, it is unnecessary and often impossible to work out the equation and sketch from there. Of course if it's a really easy one like y=x^3+2 it might be easier to just say that dy/dx = 3x^2 and do it that way, but generally it won't work out out quite as easy as that.

Do we have to find the equation of the above? And then differentiate it?

I wouldn't recommend doing this - infact, it's probably not even possible for some of them (not enough information)

The key here is to simply recognise that a derivative is simply the gradient function of the original function. So you simply look at a point, guess its gradient, and then plot it on your derivative graph, and repeat for many points.

Use turning points to guide you, as this is when the derivative is equal to zero.

Tip: It helps to start by drawing a new graph directly under the original graph, and lining up the x-values to make it easier.

I've done g for you as an example: