1. Differentiation1.1. Basics1.2. Exercise 3.6.511.3. Exercise 3.7.92. A Discontinuous Derivative1.4. Exercise 3.8.35. Implicit Differentiation1.5. Exercise 3.9.44. Related Rates1.6. Exercise 3.8.55. Lemniscate CurveMath 31A 2009.10.20MATH 31A DISCUSSIONJED YANG1. Differentiation1.1. Basics. Given a function f (x). The slope of the tangent line at x = c is f′(c).1.1.1. Higher Derivatives. Recursively define higher derivatives: f(n)= (f(n−1))′.1.1.2. Chain Rule. If f and g are differentiable, then (f ◦g)(x) = f (g(x)) is differ-entiable and (f(g(x))′= f′(g(x))g′(x).1.2. Exercise 3.6.51. Show that a nonzero polynomial function y = f(x) cannotsatisfy the equation y′′= −y. Use this to prove that neither sin x nor cos x is apolynomial.Proof. Let y = anxn+ an−1xn−1+ . . . + a1x + a0, with an6= 0. If the degree n < 2,then y′′≡ 0 so y′′6= −y. Otherwise, y′= nanxn−1+(n−1)an−1xn−2+. . .+2a2x+a1, and y′′= (n −1)nanxn−2+ . . . + 2a2. Since y′′lacks a monomial xn, we cannothave y′′= −y. 1.3. Exercise 3.7.9 2. A Discontinuous Derivative. Use the limit definition toshow that g′(0) exists but g′(0) 6= limx→0g′(x), whereg(x) =x2sin1xif x 6= 0,0 if x = 0.Proof. Recall g′(x) = limh→0g(x+h)−g(x)h. So by definition, g′(0) = limh→0g(h)−g(0)h=limh→0h sin1h. Using Squeeze Theorem and −|h| ≤ h sin1h≤ |h|, we get thatg′(0) = 0. Away from x = 0, we can use the formula and get g′(x) = 2x sin1x+x2cos1x·(−1)1x2= 2x sin1x−cos1x. Now limx→0g′(x) does not exist since limx→02x sin1x=0 by Squeeze Theorem but limx→0cos1xdoes not exist. 1.4. Exercise 3.8.35. Implicit Diff erentiation. If the deriva tive dx/dy existsat a point and dx/dy = 0, then the tangent line is vertical. Calculate dx/dy for theequation y4+ 1 = y2+ x2and find the points on the graph where the tangent lineis vertical.Solution. By implicit differentiation, we get 4y3= 2y + 2xdxdy. Setting dx/dy = 0,we get 4y3= 2y, so y = 0, ±1√2. I f y = 0 then x = ±1, if y = ±1√2, then x2=34,so we get 6 po ints. Math 31A Yang 21.5. Exercise 3.9.44. Related Rates. A wheel of radius r is centred at theorigin. As it rotates, the rod of length L attached at the point P drives a pistonback and forth in a straight line. Le t x be the distance from the origin to the pointQ at the end of the rod.(a) Use the Pythagorean Theorem to show thatL2= (x −r cos θ)2+ r2sin2θ.(b) Differentiate part (a) with resepct to t to prove that2(x − r cos θ)dxdt+ r sin θdθdt+ 2r2sin θ cos θdθdt= 0.(c) Calculate the speed of the piston when θ =π2, assuming that r = 10 cm,L = 30 cm, and the wheel rotates at 4 revolutions per minute.Solution. Parts (a) and (b) are str aightforward. 4 revolutions per minute meansdθdt= 4 · 2π per minute. From part (a), we get 302= x2+ 102, so x = 20√2.Plugging in, we get 2(20√2 − 0)(dxdt+ 10 · 8π) + 0 = 0. Sodxdt= −80π cm perminute. 1.6. Exercise 3.8.55. Lemniscate Curve. The lemniscate curve (x2+ y2)2=4(x2−y2) was discovered by Jacob Bernoulli in 1694, who noted that it is “shapedlike a figure 8, or knot, or the bow of a ribbo n.” Find the coordinates of the fourpoints at which the tangent line is horizontal.Solution. By implicit differentiation and chain rule, we get 2(x2+ y2)(2x + 2yy′) =4(2x − 2yy′). If y′= 0, we get 2(x2+ y2)(2x) = 4(2x), yielding x2+ y2= 2.Substituting in to the lemniscate curve, we get x2− y2= 1. So x2= 3/2 andy2= 1/2.