Rules of Differentiation. Lecture 12. Product and Quotient Rules.

Transcription

1 Rules of Differentiation. Lecture 12. Prouct an Quotient Rules. We warne earlier that we can not calculate the erivative of a prouct as the prouct of the erivatives. It is easy to see that this is so. Inee the simple function f(x) = x 2 is the prouct of the two (equal) functions g(x) = x an h(x) = x. But the prouct of the erivatives of g(x) an h(x) is g (x)h (x) = 1 1 = 1, not even close to the erivative (= slope) of the parabola f(x) = x 2. So we nee some other way to compute erivatives of proucts an quotients. Fortunately, this is fairly easy. One quick example illustrates nicely what shoul an oes work: Example 1. Suppose the sies of a rectangle are both changing length. If we know how fast the sies are changing, can we fin how fast the area is changing? Let s set this up an see what this has to o with erivatives of proucts. Represent the length of the rectangle by l an the with by w, an suppose that both l an w are changing as functions of time. The area of the rectangle A = lw is then also changing as a function of time. To etermine how fast the area A is changing we first calculate the average rate of change of the area with time. But if t is some change in time, an l an w are the corresponing changes in l an w, then the change in A is

2 w l w l l w l w A = lw + l w + w l + l w lw = l w + w l + l w so the average rate of change of the area is A t l w + w l + l w = t = l w t + w l t + l w t To get the instantaneous rate of change of the area we take the limit of this average rate of change as the time interval t goes to zero. Note, though, that as t 0, the tiny area ( l)( w) vanishes, so that that is, A t = t (lw) = ( lim l w t 0 t + w l t + l w ) t = l w t + wl t ; (lw) = lw t t + wl t. 2

3 Then the general formal rules for calculating erivatives of proucts an quotients are just The Prouct Rule. x (fg) = ( f)g + f( x x g), or (fg) = f g + fg. The Quotient Rule. or ( ) f x g = ( f)g f( x x g), g 2 ( ) f = f g fg. g g 2 A proof of the Prouct Rule is not particularly ifficult, but it is sufficiently time consuming that we shall omit it from class iscussion. However, we o suggest that you check out the proof of the Prouct Rule in the text. The Quotient Rule is just a ifferent version of the Prouct Rule. In fact, here is how you can quickly erive the Quotient Rule shoul you forget it! 3

4 Example 2. Consier the two functions f(x) = x 2 an g(x) = x 3. Then their prouct is the function f(x)g(x) = x 5, with erivative x (fg) = x x5 = 5x 4. Calculating this erivative using the Prouct Rule gives us x (fg) = (f (x))g(x) + f(x)(g (x)) = (2x)x 3 + (x 2 )(3x 2 ) = 5x 4, in complete agreement with the previous calculation! Example 3. Let s try a more serious example. Differentiate f(x) = (2x 3 1)(2 + 4x 5 ). Of course, one thing we coul o is to multiply it out an then ifferentiate using our earlier rules. But this is the prouct of two functions, f(x) = g(x)h(x) where g(x) = 2x 3 1 an h(x) = 2 + 4x 5, so let s use the prouct rule: f (x) = = g (x)h(x) + g(x)h (x) = (2x 3 1) (2 + 4x 5 ) + (2x 3 1)(2 + 4x 5 ) = (6x 2 )(2 + 4x 5 ) + (2x 3 1)(20x 4 ) = 64x 7 20x x 2 4

5 Example 4. Suppose that f, g, an h are functions of x with Then from the Prouct Rule, or f = g, or equivalently g = fh. h g = f h + fh = f h + g h h, f = (g g h h )/h = g h gh h 2. Example 5. Fin the slope of the line tangent to y = f(x) at x = 3 where f(x) = 5x + 1 x 2 1. This calls for the Quotient Rule. So with g(x) = 5x + 1 an h(x) = x 2 1 an f(x) = g(x) h(x), we have f (x) = h(x)g (x) h (x)g(x) (h(x)) 2 = (x2 1)(5) (2x)(5x + 1) (x 2 1) 2 = 5x2 + 2x + 5 (x 2 1) 2 Then f (3) = 7/8 an f(3) = 2, so the line tangent to y = f(x) at (3, 2) is y = 7 8 (x 3) + 2 = 7 8 x Example 6. Before we leave this topic, let s ask how to eal with a prouct of more than two functions, say Now, how o we calculate f (x)? f(x) = g(x)h(x)k(x). 5

6 The Chain Rule. There is yet another class of functions that we must be prepare to eal with, the composite functions; these are functions of functions. For instance if y = f(x) is a function of a variable x an x = g(t) is a function of a variable t, then y = f(g(t)) is a composite function of t. Here is a simple example that illustrates the notion of a composite function an how we eal with their erivatives. Example 1. A ship sinks an spills a great eal of oil that forms a growing circular slick on the surface. Suppose the raius r of the slick is changing with time, say r = r(t). But the area of the slick A = πr 2 is a function of the raius an hence, it too is a function of time; that is the area is a composite function of time, A(t) = π(r(t)) 2. So the question is if we know how fast the raius is changing then how fast is the area changing? Well, in this case we can use the Prouct Rule, t A(t) = t (π(r r) = π (r r) t ( = π (rr + r r) = π 2r r ) t Or in a more suggestive form A A(t) = t t = A r r t = 2πrr t. = 2πr r t. 6

7 The Chain Rule. If f is a ifferentiable function of u an u is a ifferentiable function of x, then f(x) = f(u(x)) is a ifferentiable function of x with x f(u) = f (u) u x. In alternate notation f x = f u u x. We will forego a formal proof of the Chain Rule. However, as we shall see, it is certainly consistent with the other rules we shall use. Example 2. Consier f(x) = (x 2 + 1) 5. With enough stamina we coul multiply this out an ifferentiate it using the Aition an Power Rules or even the Prouct Rule. But it is far easier to observe that f is a simple function of the variable u(x) = x So f(x) is the composite function So using the Chain Rule f x = f u f(u) = u 5 where u(x) = x u x = (5u 4 )(2x) = 5(x 2 + 1) 4 (2x) = 10x(x 2 + 1) 4. Applying the Chain Rule is sort of like peeling an onion. Each time you remove a layer you ifferentiate with respect to the next layer, then ifferentiate the next layer with respect to its next layer. You keep this up as long as there are layers an finish by multiplying the results! 7

8 Example 3. Consier a triple ecker f(x) = ( x ) 4. The biggest problem here is to issect this function. Starting from the outsie we have f(u) = u 4 where u = x , an then So an u = v 5 where v = x f x = f u v u v x f(x) = f(u(v(x))) = (4u 3 )( 1 2 v 1/2 )(2x) = 4( x ) 3 2(x 2 + 1) 1/2 (2x) = 16x ( x ) 3. x2 + 1 Whew!! 8