Instruction
1
If the graph of the derivative should be straight and parallel to the axis OX, the equation Y' = k, then the sought-for function Y = k*x. If the graph of the derivative is a straight line passing at an angle to the numeric-axis, the graph of the parabola. If the graph of the derivative is similar to a hyperbole, even before his research, we can assume that the integral is a function of the natural logarithm. If the graph of the derivative — sine wave, the function is the cosine of the argument.
2
If the graph of the derivative is a straight line, then its equation in General form we can write Y'=k*x+b. To determine the coefficient k in a variable x will be a parallel schedule straight line through the origin. Remove this auxiliary graph of the x and y coordinates of an arbitrary point and calculate k= y/x. Sign k install in the direction of the graph of the derivative — if with the increase in the value of the argument graph rises, hence k>0. The value of the intercept b is equal to the value of Y' when x=0.
3
Determine the formula of a function compiled the equation of the derivative:
Y=k/2 * x2+bx+C

Free member with to find the schedule of the derivative is impossible. The position of the function along the Y-axis is not fixed. Points plot the resulting function is a parabola. The branches of the parabola are directed upwards if k>0 and down if k<0.
4
The graph of the derivative of exponential function coincides with the graph of the function, since the differentiation of the exponential function does not change. The control point on the graph has coordinates (0, 1), since any number to the power of zero equal to one.
5
If the graph of the derivative is the hyperbola with branches in the first and third quarters of the coordinate axes, the equation of the derivative Y' = 1/x. Therefore the integral will be a function of the natural logarithm. The reference point when plotting the function (1,0) and (e, 1).