Derivatives an the Prouct Rule James K. Peterson Department of Biological Sciences an Department of Mathematical Sciences Clemson University January 28, 2014 Outline Differentiability Simple Derivatives The Prouct Rule
This lecture will go over the etails of erivatives an the prouct rule. Lots of new etails an ieas, so pull up a chair an grab your favorite beverage. We have alreay iscusse what a erivative means in our opening salvo on limits. I m sure you remember the pain an the agony. Still, we o nee to get proper tools to buil our moels, so let s revisit this again but this time with a bit more etail. Differentiability is an iea we nee to iscuss more carefully. From what we have sai before, it seems a formal efinition of a erivative at a point woul be this.
Definition Differentiability of A Function At A Point: f is sai to be ifferentiable at a point p in its omain if the f (t) f (p) limit as t approaches p, t p, of the quotients t p exists. When this limit exists, the value of this limit is enote by a number of possible symbols: f (p) or f t (p). This can also be phrase in terms of the right an left han limits f (p + f (t) f (p) ) = lim t p + t p an f (p f (t) f (p) ) = lim t p t p. If both exist an match at p, then f (p) exists an the value of the erivative is the common value. All of our usual machinery about limits can be brought to bear here. For example, all of the limit stuff coul be rephrase in the ɛ δ framework but we selom nee to go that eep. However, the most useful way of all to view the erivative is to use an error term. Let E(x p) = f (x) f (p) f (p) (x p).
If the erivative exists, we can use the ɛ δ formlism to get some important information about the error. Choose ɛ = 1. Then by efinition, there is a raius δ so that x p < δ = f (x) f (p) f (p) x p < 1. We can rewrite this by getting a common enominator as x p < δ = f (x) f (p) f (p)(x p) x p < 1. But the numerator here is the error, so we have x p < δ = E(x p) x p < 1. This tells us x p < δ = E(x p) < x p. So as x p, x p 0 an the above tells us E(x p) 0 as well. Goo to know as we will put this fact to use right away.
Definition Error Form for Differentiability of A Function At A Point: Let the value of the erivative of f at p be enote by f (p) an let the error term E(x p) be efine by E(x p) = f (x) f (p) f (p) (x p). Then if f (p) exists, the arguments above show us E(0) = 0 (i.e x = p an everything isappears). E(x p) 0 as x p an E(x p)/(x p) 0 as x p also. So the error acts like (x p) 2! Going the other way, if there is a number L so that E(x p) = f (x) f (p) L (x p) satisfies the same two conitions, then f has a erivative at p whose value is L. We say f has erivative f (p) if an only if the error E(x p) an E(x p)/(x p) goes to 0 as x p. Here is an example which shoul help. We will take our ol frien f (x) = x 2. Let s look at the erivative of f at the point x. We have E( x) = f (x + x) f (x) f (x) x = (x + x) 2 x 2 2x x = 2x x + ( x) 2 2x x = ( x) 2. See how E( x) = ( x) 2 0 an E( x)/ x = x 0 as x 0? This is the essential nature of the erivative. Replacing the original function value f (x + x) by the value given by the straight line f (x) + f (x) x makes an error that is roughly proportional to ( x) 2. Goo to know.
Also, note using the error efinition, if we can fin a number L so that E(x p) = f (x) f (p) L (x p) goes to zero in the two ways above, we know f (p) = L. This is a fabulous tool. A funamental consequence of the existence of a erivative of a function at a point t is that it must also be continuous there. This is easy to see using the error form of the erivative. If f has a erivative at p, we know that f (x) = f (p) + f (p) (x p) + E(x p) as x p, we get lim x p f (x) = f (p) because the other terms vanish. So having a erivative implies continuity. Think of continuity as a first level of smoothness an having a erivative as the secon level of smoothness. So things at the secon level shoul get the first level for free! An they o. Theorem Differentiability Implies Continuity Let f be a function which is ifferentiable at a point t in its omain. Then f is also continuous at t. Proof We just i this argument!
We can show the erivative of f (x) = x 3 is 3x 2. Using this, write own the efinition of the erivative at x = 1 an also the error form at x = 1. State the two conitions on the error too. The efinition of the erivative is The error form is y (1) = lim x h 0 (1 + h) 3 (1) 3. h x 3 = (1) 3 + 3 (1) 2 (x 1) + E(x 1) where E(x 1) an E(x 1)/(x 1) both go to zero as x 1. We can show the erivative of f (x) = sin(x) is cos(x). Using this, write own the efinition of the erivative at x = 2 an also the error form at x = 2. State the two conitions on the error too. The efinition of the erivative is The error form is y (2) = lim x h 0 sin(2 + h) sin(2). h sin(x) = sin(2) + cos(2) (x 2) + E(x 2) where E(x 2) an E(x 2)/(x 2) both go to zero as x 2.
We know f (x) = x 5 has a erivative at each x an equals 5x 4. Explain why f (x) = x 5 must be a continuous function. x 5 is continuous since it has a erivative. Suppose a function has a jump at the point x = 5. Can this function have a erivative there? No. If the function i have a erivative there, it woul have to be continuous there which it is not since it has a jump at that point. Homework 12 12.1 Suppose a function has a jump at the point x = 2. Can this function have a erivative there? 12.2 We know f (x) = cos(x) has a erivative at each x an equals sin(x). Explain why f (x) = cos(x) must be a continuous function. 12.3 We can show the erivative of f (x) = x 7 is 7x 6. Using this, write own the efinition of the erivative at x = 1 an also the error form at x = 1. State the two conitions on the error too. 12.4 We can show the erivative of f (x) = x 2 + 5 is 2x. Using this, write own the efinition of the erivative at x = 4 an also the error form at x = 4. State the two conitions on the error too.
We nee some fast ways to calculate these erivatives. Let s start with constant functions. These never change an since erivatives are suppose to give rates of change, we woul expect this to be zero. Here is the argument. Let f (x) = 5 for all x. Then to fin the erivative at any x, we calculate this limit y f (x + h) f (x) (x) = lim x h 0 h 5 5 = lim h 0 h = lim 0 = 0. h 0 A little thought shows that the value 5 oesn t matter. So we have a general result which we ignify by calling it a theorem just because we can! Theorem The Derivative of a constant: If c is any number, then the function f (t) = c gives a constant function. The erivative of c with respect to t is then zero. Proof We just hammere this out!
Let s o one more, the erivative of f (x) = x. This one is easy too. We calculate y f (x + h) f (x) (x) = lim x h 0 h x + h x = lim h 0 h h = lim h 0 h = lim 1 = 1. h 0 So now we know that x (x) = 1 an it wasn t that har. To fin the erivatives of more powers of x, we are going to fin an easy way. The easy way is to prove a general rule an then apply it to the two examples we know. This general rule is calle the Prouct Rule. Theorem The Prouct Rule: If f an g are both ifferentiable at a point x, then the prouct fg is also ifferentiable at x an has the value ( f (x) g(x)) = f (x) g(x) + f (x) g (x) Proof We nee to fin this limit at the point x: (fg)(x) = lim x h 0 f (x + h) g(x + h) f (x) g(x). h
Proof Looks forbiing oesn t it? Let s a an subtract just the right term: x (fg)(x) = f (x + h)g(x + h) f (x) g(x + h) + f (x) g(x + h) f (x) g(x) lim. h 0 h Now group the pieces like so: x (fg)(x) = ( ) ( ) f (x + h)g(x + h) f (x)g(x + h) + f (x)g(x + h) f (x)g(x) lim. h 0 h Proof Factor out common terms: x (fg)(x) = ( ) ( ) f (x + h) f (x) g(x + h) + g(x + h) g(x) f (x) lim. h 0 h Now rewrite as two separate limits: ( ) f (x + h) f (x) (fg)(x) = lim g(x + h) x h 0 h ( ) g(x + h) g(x) + lim h 0 h f (x)
Proof Almost there. In the first limit, the first part goes to f (x) an the secon part goes to g(x) because since g has a erivative at x, g is continuous at x. In the secon limit, the f (x) oesn t change an the other piece goes to g (x). So there you have it: x (f (x) g(x)) = f (x) g(x) + f (x) g (x). We are now reay to blow your min as Jack Black woul say. Let f (x) = x 2. Let s apply the prouct rule. The first function is x an the secon one is x also. x (x 2 ) = x (x) x + (x) x (x) = (1) (x) + (x) (1) = 2 x. This is just what we ha before! Let f (x) = x 3. Let s apply the prouct rule here. The first function is x 2 an the secon one is x. x (x 3 ) = x (x 2 ) x + (x 2 ) x (x) = (2 x) (x) + (x 2 ) (1) = 3 x 2. Let f (x) = x 4. Let s apply the prouct rule again. The first function is x 3 an the secon one is x. x (x 4 ) = x (x 3 ) x + (x 3 ) x (x) = (3 x 2 ) (x) + (x 3 ) (1) = 4 x 3.
We coul go on, but you probably see the pattern. If P is a positive integer, then x x P = P x P 1. That s all there is too it. Just to annoy you, we ll state it as a Theorem: Theorem The Simple Power Rule: Positive Integers: If f is the function x P for any positive integer P, then the erivative of f with respect to x satisfies (x P) = P x P 1 Proof We just reasone this out! Next, we on t want to overloa you with a lot of teious proofs of various properties, so let s just get to the chase. From the way the erivative is efine as a limit, it is pretty clear the following properties hol: first the erivative of a sum of functions is the sum of the erivatives: Secon, the erivative of a constant times a function is just constant times the erivative. x (c f (x)) = c (f (x)). x Arme with these tools, we can take a lot of erivatives!
Polynomials: Fin ( t 2 + 4 ). ( t 2 + 4) ) = 2t. Polynomials: Fin ( 3 t 4 + 8 ). ( 3 t 4 + 8 ) = 12 t 3.
Fin ( 3 t 8 + 7 t 5 2 t 2 + 18 ). ( 3 t 8 + 7 t 5 2 t 2 + 18 ) = 24 t 7 + 35 t 4 4 t. Prouct Rules: Fin ( (t 2 + 4) (5t 3 + t 2 4) ). ( (t 2 + 4) (5t 3 + t 2 4) ) = (2t) (5t 3 + t 2 4) + (t 2 + 4) (15t 2 + 2t) an, of course, this expression above can be further simplifie.
Fin ( (2t 3 + t + 5) ( 2t 6 + t 3 + 10) ). ( (2t 3 + t + 5) ( 2t 6 + t 3 + 10) ) = (6t 2 + 1) ( 2t 6 + t 3 + 10) + (2t 3 + t + 5) ( 12t 5 + 3t 2 ) Homework 13 13.1 Fin ( 6 t 7 + 4 t 2). 13.2 Fin ( 3 t 2 + 7 t + 11 ). 13.3 Fin ( 2 t 4 + 7 t 2 2 t + 1 ). 13.4 Fin ( 2 + 3 t + 7 t 10). 13.5 Fin ( 3 t + 7 t 8).
Homework 13 13.6 Fin ( (6t + 4) (4t 2 + t) ). 13.7 Fin ( (5t 4 + t 3 + 15) (4 t + t 2 + 3) ). 13.8 Fin ( ( 8t 3 + t 4 ) (4 t + t 8 ) ). 13.9 Fin ( ( 10t 7 + 14t 5 + 8t 2 + 8) (5t 2 + t 9 ) ). 13.10 Fin ( (8t 11 + t 15 ) ( 5t 2 + 8t 2) ).