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3.4 prod quot rule.pdf

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Department
Mathematics
Course
MAT237Y1
Professor
Dan Dolderman
Semester
Fall

Description
Calc 1 Fri 9/26/08 3.4 - Product and Quotient Rules Okay, so I mentioned earlier that derivative of the product of two functions is not the same thing 2 as the product of the derivatives. Here’s why not: let f(x) ▯ x a▯▯ g(x) =▯x. The derivative of d 3 2 d d 2 the product is x = 3x , but the product of the derivatives is x x = 2x. The last dx dx dx time I checked, those are not the same (well, they are if x = 0, but you know what I mean). So, what exactly happens when you take the derivative of the product? Well, let’s see, using (what else but) the de▯nition of the derivative! Product Rule: d f(x + h)g(x + h) ▯ f(x)g(x) [f(x)g(x)] = lim dx h!0 h f(x + h)g(x + h) + f(x + h)g(x) ▯ f(x + h)g(x) ▯ f(x)g(x) = lim h!0 h Whoa, what just happened there? We added zero! This is something typical that happens where after you know what to do, it’s pretty obvious that it’s correct; the hard part is knowing that you’re supposed to \add zero" there! Let’s continue so you can see why I just added and subtracted that f(x + h)g(x): ▯ ▯ f(x + h)g(x + h) ▯ f(x + h)g(x) f(x + h)g(x) ▯ f(x)g(x) = lim + h!0▯ h ▯ g(x + h) ▯ g(x) f(x + h) ▯ f(x) = lim f(x + h) + g(x) h!0 h h ▯ ▯▯ ▯ ▯ ▯▯ ▯ = lim f(x + h) lim g(x + h) ▯ g(x) + lim g(x) limf(x + h) ▯ f(x) h!0 h!0 h h!0 h!0 h d d = f(x) ▯ dx [g(x)] + g(x)dx [f(x)] 0 0 0 So, in other words, (fg) = fg + gf . Make sure you memorize this! You WILL be tested on it! Example: ▯nd dy if y = (x + 2)(5x ▯ 4x ▯ 1). dx 0 3 d 4 2 4 2 d 3 y = (x + 2) (5x ▯ 4x ▯ 1) + (5x ▯ 4x ▯ 1) (x + 2) 3 dx 3 4 2 2 dx = (x + 2)(20x ▯ 8x) + (5x ▯ 4x ▯ 1)(3x ) This is totally correct, and you can leave your answer just like this. However, I’m going to expand it out so that we can compare. 1 3 3 4 2 2 6 4 3 6 4 2 (x + 2)(20x ▯ 8x) + (5x ▯ 4x ▯ 1)(3x ) = 20x ▯ 8x + 40x ▯ 16x + 15x ▯ 12x ▯ 3x 6 4 3 2 = 35x ▯ 20x + 40x ▯ 3x ▯ 16x So, let’s now expand in the beginning and then take the derivative and see if we get the same answer (something you do if you forget the product rule. However, be careful because on this test, we may ask you to speci▯cally to take the derivative using the product rule, so if you don’t remember, you’re screwed!). 7 5 3 4 2 y = 5x ▯ 4x ▯ x + 10x ▯ 8x ▯ 2 0 6 4 3 2 y = 35x ▯ 20x + 40x ▯ 3x ▯ 16x Well, by golly! We got the same answer! Both ways are completely correct; however, many times, this second method we just used is much more tedious and usually involves a lot more busywork. Your best method is to just memorize the product rule! 0 p 3=2 2 Example: ▯nd f (x) where f(x) = x(x ▯ x + 6x). p d d p f (x) = x (x3=2▯ x + 6x) + (x3=2▯ x + 6x) x dx dx p ▯ p ▯ = x 3 x ▯ 2x + 6 + (x3=2▯ x + 6x) p1 2 2 x Again, this is totally correct. You don’t have to simplify if it doesn’t ask you to. If you would like, you can go ahead and expand ▯rst and then take the derivative, like in the previous example and see if you get the same thing. Okay, just like how we have problems with the product of two functions when we take the derivative, we also have problems when we divide them. So, let’s derive the quotient rule! Quotient Rule: ▯ ▯ f(x+h) f(x) d f(x) g(x+h)▯ g(x) = lim dx g(x) h!0 h f(x+h)g(x) f(x)g(x+h) g(x+h)g(x) g(x)g(x+h) = lim h!0 h f(x+h)g(x)▯f(x)g(x+h) g(x+h)g(x) = lim h!0 h f(x + h)g(x) ▯ f(x)g(x + h) + f(x
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