In that Lockhart
Put on the spot that recent Friday, I tried to recall my own triumphs with the Calculus in the 1960s but could only really remember that differentiation was a technique for measuring the slope of the graph of a function. I wasn't much help to my struggling students. I should maybe have forced them to do some Excel because that would at least have distracted them from their calculus-anxiety. In fact a bit of plotting with Excel helps (me) make sense of the rules of differentiation. Because I knew that the derivative of a function [which most of us get by knowing The Rules through rote learning] tells you what the slope of the graph of that function is at any place on the graph. If f(x) = 2x (or, put another way, if y = 2x) then the derivative (following the rules) dx/dy = 2. This is telling us that for all values of x the slope of the graph equal 2 or that it is a, fairly steep, straight line.
For f(x) = x2, the values of y start off large and positive, decrease stedaily and then rise again. The derivative (following the rules) dx/dy = 2x. This says that the slope of f(x) = x2 varies as a function of x. For negative values of x, the slope is always negative (below the x-axis) and the slope is always positive for positive values of x. When x = 0 the slope is zero because the curve is flat.
Furthermore, as x gets further from zero, the slope of the graph gets bigger. How much bigger? Twice the value of x bigger. When x = 0 the slope is zero or flat. When x = 0.5 the slope dx/dy = 2x is 1 [in other words the angle of the curve, which they call a parabola, at that point = 45o].
For f(x) = x3, then dx/dy = 3x2. Here the graph of the function [blue dots] always trends upwards, there is no negative slope, so all values of the derivative [orange dots] are above the zero line.
But the slope [dx/dy = 3x2] of x3 starts off steep [when x = -4, dx/dy = 48], eases off to no slope at all when x = 0, and then gets progressively steeper again. And this is shown in the high values for the derivative away from zero; but, as for the previous graph of x2, the derivative = zero [flat line] when x = zero.
For f(x) = x3, then dx/dy = 3x2. Here the graph of the function [blue dots] always trends upwards, there is no negative slope, so all values of the derivative [orange dots] are above the zero line.
But the slope [dx/dy = 3x2] of x3 starts off steep [when x = -4, dx/dy = 48], eases off to no slope at all when x = 0, and then gets progressively steeper again. And this is shown in the high values for the derivative away from zero; but, as for the previous graph of x2, the derivative = zero [flat line] when x = zero.
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