calculus sin frontera Section 2.8: The Power Chain Rule I want to see what happens when I take a function I know, like g(x) = x 2, and raise it to a power, f (x) = ( x 2 ) a : f (x) = ( x 2 ) a = ( x 2 ) 2 if a=2 x 2 if a=- x 2 if a = 2 Figure : Critical Points: New, Old, I The function f (x) = x 2 in black; y = ( x 2 ) 2 in red. Check out where the critical points are. I ll start with looking closely at an example: f (x) = ( x 2 ) 2. The basic g is g(x) = x 2, which has one critical point at x = 0. Now, where are the critical points of f? If you look at Figure, you ll see f has a critical point, marked in blue, exactly where g had its critical point. But wait: there s more! f has two more critical points, marked in green, exactly where g has intercepts. Figure 2 repeats the story, except that the critical points for f are cuspy-looking pointy places. But they re still happening at the intercepts of g. Summary In the process of going from g(x) to [g(x)] a, i) All the critical points g had, are at the sample place for g a. ii) g a gets new critical points, but they come from the intercepts of g. Figure 2: Critical Points II The function f (x) = x 2 in black; y = ( x 2 ) 3 2 in red. Note the critical point for y = x 2 coincides with a critical point for y = ( x 2 ) 2 3, the latter has two more critical points. Any formula for [g a ] will have to explain these observations, and here s a great formula, called the Power-Chain Rule: If y = [g(x)] a, Then y = a [g(x)] a g (x) or, more casually, [g a ] = ag a g. So, g a = 0 critical points occur exactly when the factors of g a are zero: g a = 0 or g = 0 if g has an intercept if g has a critical point This is pretty much what we saw in Figures and 2. Note, though: this only accounts for critical points that occur when a derivative is zero; we d have to be more careful, in dealing with the critical points arising from c where f (c) = dne.
2 kathy davis I want to work examples two of the 4U examples from the lecture. But first I want to see where the power-chain rule comes from. I ll start with some silly examples, really just algebra, starting with g 2. It s just a product rule: [ g 2] = [g g] = [g] [g] + [g] [g] = 2 [g] [g] = 2g 2 g No surprise: we can the same thing with cubes, [ g 3] = [ g g 2] = [g] [g 2] + [g] [g 2] = g g 2 + g [2gg ] = g g 2 + 2g g 2 = 3g 3 g This next one is a little surprise: [ ]. [ ] 2 g Start with truth: g = g. Now differentiate truth: [ ( g) 2] = g. But that s a square thingie: if h = g then I m just writing [ h 2] = g, and I can differentiate square thingies it s the first algebra computation I did: 2hh = g, or 2 g( g) = g. Now you just solve for ( g) : ( g) = g /2 g. If [ ] you get rid of square roots, though, you get g 2 = g /2g 2, or [ ] g 2 = 2 g 2 g = 2 g 2 g Now for the whole thing: since the rule involves g a, it can t be very surprising that I ll use Newton s Binomial Theorem: (x + h) a = x a + ax a h + a(a ) ax a 2 h 2 + a(a )(a 2) ax a 3 h 3 + 3! Then too, we re trying to compute a derivative, so we ll have to compute quantities like [g(x + h)] a [g(x)] a, and for this, I ll use a tangent line approximation. It makes perfect sense: I keep saying g behaves just like its tangent line, and here s where I actually use the tangent line approximation, g(x + h) g(x) + g (x)h We expect a (x a) factor near g (a), but x a = (x + h) x = h, so the h takes the place of the (x a). Then [g(x + h)] a [ g(x) + g (x)h ] a I ll simplify using the Binomial Theorem: [ g(x) + g (x)h ] a = [g(x)] a + a [g(x)] a [ g (x)h ] + a(a ) = [g(x)] a + a [g(x)] a [ g (x) ] h + a(a ) [g(x)] a 2 [ g (x) ] 2 h2 [g(x)] a 2 [ g (x)h ] 2
calculus sin frontera 3 Now when I compute [g(x + h)] a [g(x)] a I get a [g(x)] a [ g (x) ] h + so dividing by h will give me a(a ) [g(x)] a 2 [ g (x) ] 2 h2 [g(x + h)] a [g(x)] a h = a [g(x)] a [ g (x) ] + a(a ) [g(x)] a 2 [ g (x) ] 2 h and in the limit, I get a [g(x)] a [g (x)]: informally, [g a ] = ag a g. Never happier than when I can use two approximations in one computation.
4 kathy davis Example I This is the first of the 4U problems: differentiate, simplify, and find all critical points of f (x) = x 2 x 2. Before we get started, I want to see what freezing near the intercepts will get us. Let s start by factoring f into its intercepts, f (x) = x 2 + x x. Then the intercepts are: x =, x = 0, x =. Thus: Near x = f (x) ( ) 2 + x = 2 + x. Near x = 0 f (x) x 2 + 0 0 = x 2 Near x = f (x) () 2 + x = 2 x Figure 3 shows what we have; you ll notice f has vertcal tangents at x = ±, a clear critical point at zero, and probably two more between zero and ±. So, after differentiation and simplification, I expect the numerator to be zero at three places, and the denominator at two. Here goes: f (x) = [x 2] [ ] [ ] x 2 + x 2 x 2 = [2x] x 2 + x 2 2x 2 x 2 [ x 3 x = 2 2x ] x 2 x 2 x 2 = 2x x 2 x 3 x 2 Figure 3: Frozen Pieces The function is f (x) = x 2 x 2 ; these are the frozen pieces near x =, x = 0, x =. we re going to find all the critical points. = 2x x 2 x 2 x 3 x 2 = 2x( x2 ) x 3 x 2 = 2x 2x3 x 3 x 2 = 2x 3x3 = x(2 3x2 ) x 2 x 2 Critical points: Numerator Is Zero When the factors are zero; x = 0; 2 3x 2 = 0 or, x = 0, x = ± 2/3. Denominator Is Zero When x 2 = 0 or x 2 = 0 or x = ±. It tuned out just as Figure 3 shows; five critical points in all. The full graph is in Figure 4. Notice that the critical points x = ± 2/3 are approximately x = ±.66 ±.64, which would locate the critical points at about ±.8. You see that in Figure 4; the critical points are very close to ±! If you look very closely, you ll see they re a bit away from ±; after all, 2/3 isn t actually.64! Which direction from ±.8 should the actual critical points be? Figure 4: Full Graph The above function, with all its critical points.
calculus sin frontera 5 Example II This is the last 4U problem I gave: differentiate, simplify, and find all critical points of f (x) = x/(x + ) 2. A few preliminaries: Near x = f (x) ( )/( + x) 2. A vertical asymptote, but flipped because of the negative one. Also, because of the ( + x) 2, lim x ± f (x) = Near x = 0 f (x) x/( + 0) 2 = x. A 45 o straight line. Near x = ± f (x) x/(x) 2 asymptote, y = 0. = /x. A standard horizontal Here, it a a bit harder to guess how all these connect; we really need to find the critical points. f (x) = [x] ( + x) 2 x [ ( + x) 2] [( + x) 2 ] 2 = [] ( + x)2 x [2( + x)] ( + x) 4 Figure 5: Frozen Pieces The function is f (x) = x/(x + ) 2 ; these are the frozen pieces near x =, x = 0, x = ±. we re going to find all the critical points. = ( + x)2 2x( + x) ( + x) [( + x) 2x] ( + x) 4 = ( + x) 4 = Critical points: Numerator Is Zero When x = 0: x =. ( x) ( + x) 3 Denominator Is Zero When ( + x) 3 = 0, or, x =. But f has a vertical asymptote at x =, so this cannot be a critical point. Figure 6 shows how all the parts of the graph connect. The only critical point occurs as the y = x part near zero connects with the y = /x part near +. Figure 6: Full Graph The above function, all together.