Derivatives and the Product Rule James K. Peterson Department of Biological Sciences and Department of Mathematical Sciences Clemson University January 28, 2014
Outline 1 Differentiability 2 Simple Derivatives 3 The Product Rule
This lecture will go over the details of derivatives and the product rule. Lots of new details and ideas, so pull up a chair and grab your favorite beverage.
Differentiability We have already discussed what a derivative means in our opening salvo on limits. I m sure you remember the pain and the agony. Still, we do need to get proper tools to build our models, so let s revisit this again but this time with a bit more detail.
Differentiability We have already discussed what a derivative means in our opening salvo on limits. I m sure you remember the pain and the agony. Still, we do need to get proper tools to build our models, so let s revisit this again but this time with a bit more detail. Differentiability is an idea we need to discuss more carefully. From what we have said before, it seems a formal definition of a derivative at a point would be this.
Differentiability Definition Differentiability of A Function At A Point: f is said to be differentiable at a point p in its domain if the f (t) f (p) t p limit as t approaches p, t p, of the quotients exists. When this limit exists, the value of this limit is denoted by a number of possible symbols: f (p) or df dt (p). This can also be phrased in terms of the right and left hand limits f (p + f (t) f (p) ) = lim t p + t p and f (p ) = lim t p If both exist and match at p, then f (p) exists and the value of the derivative is the common value. f (t) f (p) t p.
Differentiability All of our usual machinery about limits can be brought to bear here. For example, all of the limit stuff could be rephrased in the ɛ δ framework but we seldom need to go that deep.
Differentiability All of our usual machinery about limits can be brought to bear here. For example, all of the limit stuff could be rephrased in the ɛ δ framework but we seldom need to go that deep. However, the most useful way of all to view the derivative is to use an error term.
Differentiability All of our usual machinery about limits can be brought to bear here. For example, all of the limit stuff could be rephrased in the ɛ δ framework but we seldom need to go that deep. However, the most useful way of all to view the derivative is to use an error term. Let E(x p) = f (x) f (p) f (p) (x p).
Differentiability If the derivative exists, we can use the ɛ δ formlism to get some important information about the error. Choose ɛ = 1. Then by definition, there is a radius δ so that x p < δ = f (x) f (p) f (p) x p < 1.
Differentiability If the derivative exists, we can use the ɛ δ formlism to get some important information about the error. Choose ɛ = 1. Then by definition, there is a radius δ so that x p < δ = f (x) f (p) f (p) x p < 1. We can rewrite this by getting a common denominator as x p < δ = f (x) f (p) f (p)(x p) x p < 1.
Differentiability If the derivative exists, we can use the ɛ δ formlism to get some important information about the error. Choose ɛ = 1. Then by definition, there is a radius δ so that x p < δ = f (x) f (p) f (p) x p < 1. We can rewrite this by getting a common denominator 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.
Differentiability This tells us x p < δ = E(x p) < x p.
Differentiability This tells us x p < δ = E(x p) < x p. So as x p, x p 0 and the above tells us E(x p) 0 as well. Good to know as we will put this fact to use right away.
Differentiability Definition Error Form for Differentiability of A Function At A Point: Let the value of the derivative of f at p be denoted by f (p) and let the error term E(x p) be defined 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 and everything disappears). E(x p) 0 as x p and 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 conditions, then f has a derivative at p whose value is L. We say f has derivative f (p) if and only if the error E(x p) and E(x p)/(x p) goes to 0 as x p.
Differentiability Here is an example which should help. We will take our old friend f (x) = x 2. Let s look at the derivative 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.
Differentiability Here is an example which should help. We will take our old friend f (x) = x 2. Let s look at the derivative 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 and E( x)/ x = x 0 as x 0? This is the essential nature of the derivative.
Differentiability Here is an example which should help. We will take our old friend f (x) = x 2. Let s look at the derivative 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 and E( x)/ x = x 0 as x 0? This is the essential nature of the derivative. 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. Good to know.
Differentiability Also, note using the error definition, if we can find 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.
Differentiability Also, note using the error definition, if we can find 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 fundamental consequence of the existence of a derivative 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 derivative. If f has a derivative at p, we know that f (x) = f (p) + f (p) (x p) + E(x p)
Differentiability Also, note using the error definition, if we can find 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 fundamental consequence of the existence of a derivative 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 derivative. If f has a derivative 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 derivative implies continuity.
Differentiability Think of continuity as a first level of smoothness and having a derivative as the second level of smoothness. So things at the second level should get the first level for free! And they do. Theorem Differentiability Implies Continuity Let f be a function which is differentiable at a point t in its domain. Then f is also continuous at t. Proof We just did this argument!
Differentiability Example Example We can show the derivative of f (x) = x 3 is 3x 2. Using this, write down the definition of the derivative at x = 1 and also the error form at x = 1. State the two conditions on the error too. Solution The definition of the derivative is The error form is dy (1) = lim dx h 0 (1 + h) 3 (1) 3. h x 3 = (1) 3 + 3 (1) 2 (x 1) + E(x 1) where E(x 1) and E(x 1)/(x 1) both go to zero as x 1.
Differentiability Example Example We can show the derivative of f (x) = sin(x) is cos(x). Using this, write down the definition of the derivative at x = 2 and also the error form at x = 2. State the two conditions on the error too. Solution The definition of the derivative is The error form is dy (2) = lim dx h 0 sin(2 + h) sin(2). h sin(x) = sin(2) + cos(2) (x 2) + E(x 2) where E(x 2) and E(x 2)/(x 2) both go to zero as x 2.
Differentiability Example Example We know f (x) = x 5 has a derivative at each x and equals 5x 4. Explain why f (x) = x 5 must be a continuous function. Solution x 5 is continuous since it has a derivative. Example Suppose a function has a jump at the point x = 5. Can this function have a derivative there? Solution No. If the function did have a derivative there, it would have to be continuous there which it is not since it has a jump at that point.
Differentiability Homework 12 12.1 Suppose a function has a jump at the point x = 2. Can this function have a derivative there? 12.2 We know f (x) = cos(x) has a derivative at each x and equals sin(x). Explain why f (x) = cos(x) must be a continuous function. 12.3 We can show the derivative of f (x) = x 7 is 7x 6. Using this, write down the definition of the derivative at x = 1 and also the error form at x = 1. State the two conditions on the error too. 12.4 We can show the derivative of f (x) = x 2 + 5 is 2x. Using this, write down the definition of the derivative at x = 4 and also the error form at x = 4. State the two conditions on the error too.
Simple Derivatives We need some fast ways to calculate these derivatives. Let s start with constant functions. These never change and since derivatives are supposed to give rates of change, we would expect this to be zero. Here is the argument.
Simple Derivatives We need some fast ways to calculate these derivatives. Let s start with constant functions. These never change and since derivatives are supposed to give rates of change, we would expect this to be zero. Here is the argument. Let f (x) = 5 for all x. Then to find the derivative at any x, we calculate this limit dy f (x + h) f (x) (x) = lim dx h 0 h 5 5 = lim h 0 h = lim 0 = 0. h 0
Simple Derivatives We need some fast ways to calculate these derivatives. Let s start with constant functions. These never change and since derivatives are supposed to give rates of change, we would expect this to be zero. Here is the argument. Let f (x) = 5 for all x. Then to find the derivative at any x, we calculate this limit dy f (x + h) f (x) (x) = lim dx h 0 h 5 5 = lim h 0 h = lim 0 = 0. h 0 A little thought shows that the value 5 doesn t matter. So we have a general result which we dignify by calling it a theorem just because we can!
Simple Derivatives Theorem The Derivative of a constant: If c is any number, then the function f (t) = c gives a constant function. The derivative of c with respect to t is then zero. Proof We just hammered this out!
Simple Derivatives Let s do one more, the derivative of f (x) = x. This one is easy too. We calculate dy (x) = lim dx h 0 f (x + h) f (x) h = lim h 0 x + h x h = lim h = lim 1 = 1. h 0 h 0 h
Simple Derivatives Let s do one more, the derivative of f (x) = x. This one is easy too. We calculate dy (x) = lim dx h 0 f (x + h) f (x) h = lim h 0 x + h x h = lim h = lim 1 = 1. h 0 h 0 h So now we know that d dx (x) = 1 and it wasn t that hard. To find the derivatives of more powers of x, we are going to find an easy way. The easy way is to prove a general rule and then apply it to the two examples we know. This general rule is called the Product Rule.
The Product Rule Theorem The Product Rule: If f and g are both differentiable at a point x, then the product fg is also differentiable at x and has the value ( f (x) g(x)) = f (x) g(x) + f (x) g (x) Proof We need to find this limit at the point x: d (fg)(x) = lim dx h 0 f (x + h) g(x + h) f (x) g(x). h
The Product Rule Proof Looks forbidding doesn t it? Let s add and subtract just the right term: d dx (fg)(x) = lim h 0 f (x + h)g(x + h) f (x) g(x + h) + f (x) g(x + h) f (x) g(x). h Now group the pieces like so: d dx (fg)(x) = ( ) f (x + h)g(x + h) f (x)g(x + h) lim h 0 h ( ) + f (x)g(x + h) f (x)g(x).
The Product Rule Proof Factor out common terms: d dx (fg)(x) = ( ) f (x + h) f (x) g(x + h) + lim h 0 h ( ) g(x + h) g(x) Now rewrite as two separate limits: ( ) d f (x + h) f (x) (fg)(x) = lim g(x + h) dx h 0 h ( ) g(x + h) g(x) + lim f (x) h 0 h f (x).
The Product Rule Proof Almost there. In the first limit, the first part goes to f (x) and the second part goes to g(x) because since g has a derivative at x, g is continuous at x. In the second limit, the f (x) doesn t change and the other piece goes to g (x). So there you have it: d dx (f (x) g(x)) = f (x) g(x) + f (x) g (x).
The Product Rule We are now ready to blow your mind as Jack Black would say. Let f (x) = x 2. Let s apply the product rule. The first function is x and the second one is x also. d dx (x 2 ) = d dx (x) x + (x) d dx (x) = (1) (x) + (x) (1) = 2 x. This is just what we had before!
The Product Rule We are now ready to blow your mind as Jack Black would say. Let f (x) = x 2. Let s apply the product rule. The first function is x and the second one is x also. d dx (x 2 ) = d dx (x) x + (x) d dx (x) = (1) (x) + (x) (1) = 2 x. This is just what we had before! Let f (x) = x 3. Let s apply the product rule here. The first function is x 2 and the second one is x. d dx (x 3 ) = d dx (x 2 ) x + (x 2 ) d dx (x) = (2 x) (x) + (x 2 ) (1) = 3 x 2.
The Product Rule We are now ready to blow your mind as Jack Black would say. Let f (x) = x 2. Let s apply the product rule. The first function is x and the second one is x also. d dx (x 2 ) = d dx (x) x + (x) d dx (x) = (1) (x) + (x) (1) = 2 x. This is just what we had before! Let f (x) = x 3. Let s apply the product rule here. The first function is x 2 and the second one is x. d dx (x 3 ) = d dx (x 2 ) x + (x 2 ) d dx (x) = (2 x) (x) + (x 2 ) (1) = 3 x 2. Let f (x) = x 4. Let s apply the product rule again. The first function is x 3 and the second one is x. d dx (x 4 ) = d dx (x 3 ) x + (x 3 ) d dx (x) = (3 x 2 ) (x) + (x 3 ) (1) = 4 x 3.
The Product Rule We could go on, but you probably see the pattern. If P is a positive integer, then d dx 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 derivative of f with respect to x satisfies (x P) = P x P 1 Proof We just reasoned this out!
The Product Rule Next, we don t want to overload you with a lot of tedious proofs of various properties, so let s just get to the chase. From the way the derivative is defined as a limit, it is pretty clear the following properties hold: first the derivative of a sum of functions is the sum of the derivatives:
The Product Rule Next, we don t want to overload you with a lot of tedious proofs of various properties, so let s just get to the chase. From the way the derivative is defined as a limit, it is pretty clear the following properties hold: first the derivative of a sum of functions is the sum of the derivatives: Second, the derivative of a constant times a function is just constant times the derivative. d dx (c f (x)) = c d (f (x)). dx Armed with these tools, we can take a lot of derivatives!
The Product Rule Example Polynomials: Example Find ( t 2 + 4 ). Solution ( t 2 + 4) ) = 2t.
The Product Rule Example Polynomials: Example Find ( 3 t 4 + 8 ). Solution ( 3 t 4 + 8 ) = 12 t 3.
The Product Rule Example Example Find ( 3 t 8 + 7 t 5 2 t 2 + 18 ). Solution ( 3 t 8 + 7 t 5 2 t 2 + 18 ) = 24 t 7 + 35 t 4 4 t.
The Product Rule Example Product Rules: Example Find ( (t 2 + 4) (5t 3 + t 2 4) ). Solution ( (t 2 + 4) (5t 3 + t 2 4) ) = (2t) (5t 3 + t 2 4) + (t 2 + 4) (15t 2 + 2t) and, of course, this expression above can be further simplified.
The Product Rule Example Example Find ( (2t 3 + t + 5) ( 2t 6 + t 3 + 10) ). Solution ( (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 )
The Product Rule Homework 13 13.1 Find ( 6 t 7 + 4 t 2). 13.2 Find ( 3 t 2 + 7 t + 11 ). 13.3 Find ( 2 t 4 + 7 t 2 2 t + 1 ). 13.4 Find ( 2 + 3 t + 7 t 10). 13.5 Find ( 3 t + 7 t 8).
The Product Rule Homework 13 13.6 Find ( (6t + 4) (4t 2 + t) ). 13.7 Find ( (5t 4 + t 3 + 15) (4 t + t 2 + 3) ). 13.8 Find ( ( 8t 3 + t 4 ) (4 t + t 8 ) ). 13.9 Find ( ( 10t 7 + 14t 5 + 8t 2 + 8) (5t 2 + t 9 ) ). 13.10 Find ( (8t 11 + t 15 ) ( 5t 2 + 8t 2) ).