The gradient of the curve at point (x,y) is given by dy/dx = [7 sqrt(x^5)] -4. where x>0. Find the equation of the curve given that the curve passes through the point 1,3.

I like this question because it can be broken down into parts, each which contain an important reminder/lesson for an A level/GCE maths student. I would start the lesson by running through the rules of indices. Write down any they can remember. Then running through basic calculus (reminder how to differentiate and integrate), probing them to recall anything that might trip them up (e.g +c).Anyway. Start the question by reminding them not to be intimidated by 7 sqrt(x^5)] ... how can they simplify it.... 7x^2.5.Then ask them, how do they get from dy/dx = to just y = something (integrate) .... y = 2x^3.5 - 4x .... if they forget the plus c, just gently remind them after they try to finish the question.Once they identify +c , ask them how to find it.... hopefully they see the answer is in the question with the (1,3) that they can now substitute..... y = 2x^2.5 - 4x + 5