Leibniz and the Chain Rule –MATH 1401 Spring 2007
(#1) I thought that you might bene…t from a historical pro…le of Gottfried Leibniz (1646-1716), co-inventor of Di¤erential Calculus. Leibniz is credited for giving calculus its name. He referred to it as in…nitesimal calculus. [Newton’s version of the derivative was called the ‡uxion. His paper was written in 1666 – not bad for a 23-year old...] Leibniz did not publish anything about calculus until 1684, but he wrote about many of his discoveries in letters and manuscripts as early as 1675. Although Leibniz had already seen Newton’swork, most experts agree that Leibniz’smathematical reasoning was certainly di¤erent than Newton’s. Neither de…ned the derivative in terms of limits, but Leibniz was the …rst to introduce the idea of in…nitesimally small quantities, dx and dy, and thus, identi…ed dy the slope of a tangent line to a curve as : Leibniz relied heavily on calculations based dx on geometry and graphs, but did not de…ne his version of the derivative as the slope of the tangent line. Leibniz is credited with discovering the Product Rule for derivatives. His “proof” is not rigorous and assumes the basic properties of limits (in this case, the limit as dx 0): ! dy dy (y + dy) y If represents the derivative, then we have = : dx dx dx [This, of course, resembles the di¤erence quotient, or slope of the secant line.]
If we wish to …nd the derivative of a product y = uv [It is assumed that the x’s are hidden inside the u and v:], then we have d (uv) (u + du)(v + dv) uv uv + u (dv) + (du) v + (du)(dv) uv = = dx dx dx u (dv) + (du) v + (du)(dv) dv du du = = u + v + dv dx dx dx dx dv The last term could also be written as (du) : dx Since dv is in…nitesimally small, the last term vanishes and we have d (uv) dv du = u + v: dx dx dx
[Newton would have written this as [uv]0 = uv0 + u0v:] (#2) Your challenge... Using Leibniz’snotation, “prove”this: du dv u v u d dx dx v = dx v2
1 (#3) The Product and Quotient Rules were “easy” for Leibniz. The Chain Rule would be more di¢ cult to “prove”, but its form using Leibniz notation is very simple. Recall that if we have the composite function y = f (g (x)) and dy Outer : y = f (u) f 0 (u) = ) du du Inner : u = g (x) g0 (x) = ; ) dx then since y is ultimately a function of x; we have
y0 (x) = [f (g (x))]0 = f 0 (g (x)) g0 (x) :
Since we must evaluate f 0 at g (x) ; we have dy = f (u) = f (g (x)) du 0 0 and thus, we see why Leibniz says that the derivative acts like a fraction: dy dy du = : dx du dx It looks like the du’sare being cancelled, but, in fact, all we are doing is multiplying derivatives per Newton. (#4) More notation...
Since y0 = f 0 (u) and we desire to evaluate f 0 (u) at u = g (x) ; a notation is sometimes added to the base of the parentheses. dy dy du = dx du dx u=g(x) Another popular equivalent notation uses the “evaluation bar”. dy dy du = dx du dx g(x)
I prefer to place evaluations in square brackets. It typically removes any possible confusion.
dy dy du = dx du dx g(x) In both notations, it is assumed that y is a function of u and that we are evaluating at u = g (x) : (#5) Parametric equations. Suppose we have moving particle and we desire to record its position with respect to time. The customary method would be to establish TWO functions of t –one to record the x-coordinate and another to record the y-coordinate. One of the exercises asks us to describe uniform circular motion. Suppose we have a particle on a unit circle centered at the origin: x2 + y2 = 1: We will start the particle at P (1; 0) and suggest the following two functions: x = x (t) = cos (t) y = y (t) = sin (t) ; 0 t 2:
2 (a) How do we know that this traces out a circle? Eliminate t: If we square both sides, we have
x2 = cos2 (t) y2 = sin2 (t) :
Now add the two equations together.
x2 + y2 = cos2 (t) + sin2 (t) = 1:
Thus, for any given value of t; the point (x (t) ; y (t)) must land on the unit circle. (b) You might recall that if t represents an angle in standard position, then the corresponding point on the unit circle is always (cos (t) ; sin (t)) : Furthermore, as t increases, the point of interest travels from P (1; 0) through all four quadrants (makes one complete revolution) and then returns to P:
y 1
Example:
0 .5 p3 1 When t = ; our particle is located at ; : 0 6 2 2 -1 -0.5 0 0 .5 1 ! x
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(c) Since the parameterization traces out the circle, is it possible to use the parameterization to …nd the slope of the tangent line at a particular time? The answer is YES. From Leibniz’sform of the Chain Rule, we have
dy dy dt m = = : tan dx dx dt Remember that Leibniz said that derivatives act like fractions. Let’stry it! dy d dx d = [sin (t)] = cos (t) = [cos (t)] = sin (t) dt dt dt dt Thus, the slope of the tangent line should be
dy cos (t) mtan = = = cot (t) : dx ( sin (t)) When t = ; we believe the correct answer is cot = p3: 6 6 Let’ssee if this agrees with the geometry.
3 The slope of the radius is
1 y2 y1 2 1 mradius = = = : x2 x1 p p3 3 2 !
We recall that the products of the slopes of perpendicular lines must be ( 1) : Thus, the slope of the tangent line to the circle (perpendicular to the radial segment) must be p3 mtan = = p3:X 1 !
Also, we can check this with the General Power form of the Chain Rule! The equation of the upper semicircle is
1=2 f (x) = 1 x2 = 1 x2 : p p3 The slope of the tangent line at x = is 2
2 1=2 1 2 1=2 2 1 1 1 x 0 = 1 x 1 x 0 = ( 2x) 2 2 p1 x2 h i x = : p1 x2 Thus, we have
p3 p3 p3 p3 2 ! 2 ! 2 ! f 0 = = = 2 2 3 1 ! p3 1 1 4 4 v r r u 2 ! u t p3 2 ! = = p3: 1 X 2 (#6) Notice that the slope of the tangent line does NOT depend on the parameterization. This is a di¤erent parameterization of the circle. We can still square both sides, add the equations, and we will again obtain x2 + y2 = 1:
x = cos t2 2 y = sin t p3 1 So we should see that when t = ; the particle will be located at ; : 6 2 2 r !
4 Will Leibniz give us the same slope for the tangent line? YES! Let’stry the formula again. We will invoke the Chain Rule forms for the trig. functions!
dy 2 2 2 = cos t t 0 = 2t cos t dt dx 2 2 2 = sin t t 0 = 2t sin t dt dy dy dt 2t cos t2 = = = cot t2 : dx dx ( 2t sin (t2)) dt When t = ; the slope of the tangent line must be 6 r 2 cot = cot = p3:X 6 ! 6 r (#7) Leibniz also gives us a formula for instantaneous speed along the parametric curve. Typically, if speed is a function of t; the name of the function is c (t) [since c; the constant, is the speed of light]. This text does not give the speed function a speci…c name. Here’sthe formula:
2 2 dx dy 2 2 c (t) = + = (x0 (t)) + (y0 (t)) : s dt dt q In this formula, the parameterization DOES matter. Let’s look at our original parameterization for uniform circular motion. Why is it called “uniform”? We will answer this shortly.
x = cos (t) y = sin (t)
dx 2 dy 2 c (t) = + = ( sin (t))2 + (cos (t))2 s dt dt q = sin2 (t) + cos2 (t) = 1: q If t is measured in seconds, then the speed of the particular travelling around the unit circle must be one unit per second. Since the circumference of that circle is 2r = 2 (1) = 2 units, then we see that one revolution must take exactly 2 seconds. Hence, the motion on the circle is uniform in the sense that the particle is travelling at a constant speed and that we can track its progress with the associated central angle, which is also increasing uniformly with time.
5 If we examine the alternate parameterization, we have
x = cos t2 2 y = sin t dx 2 dy 2 c (t) = + = ( 2t sin (t2))2 + (2t cos (t2))2 s dt dt q = 4t2 sin (t2) + 4t2 cos (t2) = p4t2 sin (t2) + cos (t2) p p = 2t (1) = 2t:
If assume that t 0; then p4t2 = 2t = 2t: j j This tells us that the speed increases linearly with time. At …rst, we move slowly, and then after a few more seconds, our angular speed is quite large. Here’sa graph of the x-coordinate vs. time:
x 1
0.5 We complete one revolution each time the graph achieves its 0 maximum value (1) : 0 2.5 5 7.5 10
t
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(#8) We appeal to the Leibniz form of the Chain Rule when dealing with inverses. Recall that if g is the inverse of f; then
f (g (x)) = x:
By Leibniz, we have d d [f (g (x))] = [x] dx dx df du = 1: du dx du We recall that g (x) = : 0 dx By division, we have du 1 1 = = : dx df f 0 (g (x)) du
6 Example: Let f (u) = eu: We know its inverse is g (x) = ln (x) : From this theorem, we have du 1 = g0 (x) = [ln (x)]0 = dx f 0 (g (x))
u We know f 0 (u) = e and when we substitute in u = ln (x) ; we have 1 1 [ln (x)]0 = = : eln(x) x
Let’ssee if this is reasonable. On the curve, f (x) = ex; we select a point P (ln (2) ; 2) : The slope of the tangent line is
x ln(2) f 0 (x) = e f 0 (ln (2)) = e = 2: ) The corresponding point on the inverse curve [g (x) = ln (x)] must be Q (2; ln (2)) and we recall that since we’re exchanging x for y and vice versa, that the slope of the tangent line there must be 1 1 the reciprocal, m = : Thus, it seems logical that g (2) = : tan 2 0 2
1 Later, we will use this same technique to evaluate sin (x) 0 and others.
7 (#9) Solution to the Leibniz Challenge: The Quotient Rule.
u + du u uv + v (du) uv u (dv) v (du) u (dv) u d v (v + dv) v (v + dv) v = v + dv v = = dx dx dx dx du dv v u v (du) u (dv) 1 dx dx = = v (v + dv) dx v (v + dv) When dv is in…nitesimally small, v (v + dv) becomes v2:
8