Differential calculus
Differential Calculus - Derivatives and differentials The derivative of a function
Tangent to a curve Determining the lines tangent to the graph of a function from a point outside the function
Determining the lines tangent to the graph of a function from a point outside the function, example
The derivative of a function
The slope or the gradient of the secant line joining points (x, f (x)) and (x + Dx, f (x + Dx)) given by is called the difference quotient, and represents the average rate of change of the function y = f (x). Note that in the above equation x is the given fixed point and the only variable quantity is the distance
increment, Dx.
Definition of the derivative of a function
As the distance Dx gets smaller, the average rate of change becomes more accurate, that is the secant line becomes more and more like the tangent line of the curve at the point (x, f (x)), as shows the figure above.
Therefore, the limit of the difference quotient as  Dx ® 0 that equals the slope of the line tangent to the curve at the point (x, f (x)), represents the instantaneous rate of change of the function with respect to x, or when written as where Dx is substituted with h, is called the derivative of a function f at a given point x.
Geometrically, the derivative is the slope m (gradient) of the tangent at the point (x, y) to the curve y = f (x).
 There are other common notations for the derivative like, Determining the lines tangent to the graph of a function from a point outside the function
Lines tangent to the graph of a function y = f (x) from a given point (x1, y1) outside the function are defined by two points they pass through, the given point (x1, y1) and the point of tangency (x0, y0).
To find the coordinates x0 and y0 of the points of tangency we should solve the following two equations,
(1)   y0 = f (x0)
(2)   y1 -  y0 =  f ' (x0) · (x1 - x0)
as the system of the two equations with the two unknowns x0 and y0.
The first equation states that the point of tangency must lie on the given function while, at the same time, its coordinates x0 and y0 must satisfy the equation of the tangent line (2), since the point of tangency is the common point of the tangent and the curve.
Thus for example, if given is the quadratic function  f (x) =  ax2 + bx + c and the point P1(x1, y1) then, applying the above system of equations we can calculate the coordinates of the points of tangency P0(x0, y0)  directly from the formulas, Example:   Find equations of the lines tangent to the function  f (x) = - x2 + 2x + drawn from the point (3, 2).

 Solution:  To find the coordinates x0 and y0 of the points of tangency we should solve the system of equations. Substitute given quantities, the point (3, 2) and  f (x) = - x2 + 2x + 4  into equations, Let calculate the slope  f ' (x0) of the tangent line, the equation (2). Then substitute the slope and the y0 into second equation to get the points of tangency, Therefore, the points of tangency are (2, 4) and (4, -4). To write the equations of tangents t1 and t2  we calculate their slopes m1 and m2 from f ' (x0) = - 2x0 + 2. Thus,     m1 = f ' (2) = - 2 · 2 + 2 = - 2,    m1 = - 2, and      m2 = f ' (4) = - 2 · 4 + 2 = - 6,    m2 = - 6. The tangent t1 through the point (2, 4) and m1 = - 2, and the tangent t2 through the point (4, -4) and m2 = - 6, t1 ::   y = - 2x + 8  and  t2 ::   y = - 6x + 20. Check the coordinates of the points of tangency by plugging the corresponding parameters into the above formulas.   Calculus contents C 