I read the derivative as (3/2)x.
The red line in the picture is given by the equation y=(3/2)x. This is the derivative of another function which is defined as the area between the line and the x-axis. This other function can be called the primitive of the derivative, or, more usually, the anti-derivative or integral. The area under the line is the area of a triangle. First consider a definite integral, or anti-derivative, which means the triangle has a definite, or defined area, between two limits. The blue vertical line is a vertical limit for x=2 and the green line for x=6. The corresponding y values are B=3 and A=9.
The area of triangle OBC=½OC.BC=3 square units. And the area of OAD=½OD.AD=27 square units. If we have a general point x, the area will be ½x(3/2)x=¾x². This is still a definite integral because we defined the interval as [0,x]. But if we place no limits on x so x is in the interval (-∞,∞), the integral, or anti-derivative is ¾x² plus some unspecified value which we can call c (square units). This takes into account the definite areas of the triangles calculated earlier. Just as the area of OAD was bigger than that of OBC by a certain amount (shown graphically as the area between the two triangles=27-3=24 square units) so an indefinite area is bigger than a definite area by a certain amount. This justifies saying that the anti-derivative is ¾x²+c. Only when limits can be imposed can we find c.