The curve has the equation y= (x^3)/(2x-1). Find dy/dx.
To differentiate a function with respect to (wrt) x you multiply the function by the power and then lower the power by one. For example, if f(x)=x^3, then to differentiate you multiply the function by 3 (the power) and then lower the power by 1 (3-1=2). Therefore the derivative of x^3 is 3x^2. In the question it is exactly the same however we add a few other bits in. Due to the function is a fraction can use the quotient rule (which is given in your formula books), it states when y=u/x, then dy/dx=(vdu/dx -udv/dx)/v^2. In the question, we label x^3 =u and 2x-1=v.
So from before we can say that du/dx=3x^2, and dv/dx=2. Now we can plug this into the equation. so dy/dx=({2x-1}{3x^2} - 2{x^3})/(2x-1)^2. Now we can try to simplify the equation, leaving us with (4x^3 - 3x^2)/(2x-1)^2.
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