Calculus 220, section 1.3 The Derivative notes by Tim Pilachowski We now take the idea of section 1.2 (the slope of a curve at a point P) and treat it more formally. Given a function f(x), the first derivative of f is a formula which provides the slope of the curve at any point on f. Note that the derivative of a function is itself a function. The process of finding a derivative is called differentiation. The first derivative of f has several notations: ( ) ( )[ ]xfdxddxdyxff and ,,,′′. Example A: Given ()283+−= xxxf, estimate the derivative evaluated at 322−=x, i.e. −′322f. The pictures below show a series of secant lines connecting the point −−322,322 f to other points on f. As the second point approaches −−322,322 f , the secant lines approach the tangent line. If we continued to zoom in, the secant lines would come closer and closer to horizontal. So our estimate for −′322f is 0. Example A extended: Given ()283+−= xxxf, estimate −′322f. While the visual sequence above seems to point to a particular result, it is far from accurate. A point misplaced by even a little bit will change the picture greatly. We can use decimal approximations to gain a little more precision. In the table below h will indicate the amount added to 322−=x to place the second point on the curve of f. The slope of the secant line thus formed is calculated using the familiar linear slope formula: hfhfhfhfxym−−+−=−−+−−−+−=∆∆=322322322322322322 ()≅− 322f h = ≅+− h322 ()≅+− hf 322 secant slope ≅ 10.709296863 1 -0.632993162 6.810317378 -3.898979486 10.709296863 0.1 -1.532993162 10.661307068 -0.479897949 10.709296863 0.01 -1.622993162 10.708807965 -0.048889795 10.709296863 0.001 -1.631993162 10.709291965 -0.004897979 10.709296863 0.0001 -1.632893162 10.709296814 -0.000489888 10.709296863 0.00001 -1.632983162 10.709296863 -0.000048990 10.709296863 0.000001 -1.632992162 10.709296863 -0.000004899 10.709296863 0.0000001 -1.632993062 10.709296863 -0.000000497 10.709296863 0.00000001 -1.632993152 10.709296863 0.000000000So, with an accuracy to nine decimal places, as the second point approaches −−322,322 f , the slope of the secant lines approaches 0. So our estimate for −′322f is 0. Neither of the above methods is sufficient mathematically, because even with the added precision of decimals, we cannot be sure that our answer is exact, as opposed to being extremely close. Also, neither method is practical to reproduce for every point on the curve. However, notice first of all that the slope of each sequential secant line is the difference quotient ()()hxfhxf−+. As h approaches 0, sequence of secant lines approaches the tangent line, and the sequence of slopes approaches the slope of the tangent. Example A one last time: Given ()283+−= xxxf, derive a formula for ()xf′then calculate −′322f. Answers: In mathematical notation: Given ()283+−= xxxf, ()()83lim20−=−+→xhxfhxfh; 0 Example B: Given ( )()22 find , xdxdxxf=. Answer: 2x We can go through a similar process for any function of the form ()rxxf = where r could be any number. Specifically, we’d find that ()1−=′rxrxf. Such a function is called a power function, and this property of derivatives is called the power rule.Example C: Given y = x, find dxdy. Answer: 1 Note that the answer to Example C fits with what we already know about lines in slope-intercept form. Example D: Given y = 5, find ()xf′. Answer: 0 Note that the answer to Example D fits with what we already know about horizontal lines. Example E: Given ()xxf = , find ()xf′. Answer: x21 Example F: Given ( )31xxf = , find ()xf′. Answer: 43x− Example G: Given ( )23xxxf =, find ()xf′. Answer: 3835x− Some final notions about derivatives and differentiability: In line with the concept of slope of the curve at a point being equal to the slope of the tangent line at that point, ()0>′xf implies that f is increasing. Likewise ()0<′xf implies that f is decreasing. However, the implication does not go the other way. The fact that a function is increasing does not imply that ()0>′xf. The simplest example is 3xy =. Although 3x is increasing everywhere on its domain, its derivative is not always positive. The curve momentarily “levels out” precisely when x = 0, then continues upward. At this point ()03=xdxd. A similar limitation exists for decreasing functions. Another important concept for differentiability of functions is the idea of being continuous. Specifically, the existence of a derivative at a point implies that the function is continuous at that point, i.e. its graph is not disconnected pieces. However, the implication does not go the other way. A function may be continuous, but may not be differentiable at all points. Two examples are shown below. On the left, ( )225 xxf −= is defined and has values at 5 and 5=−=xx , but the derivative does not exist since the tangent is a vertical line whose slope is undefined. On the right, ()5+−= xxf is defined and continuous for all real numbers, but the derivative does not exist at x = 0 because the tangent line on the left has a positive slope, while the tangent line on the right has a negative