Derivatives Math 120 Calculus I D Joyce, Fall 2013

Size: px

Start display at page:

Download "Derivatives Math 120 Calculus I D Joyce, Fall 2013"

Bennett Austin
7 years ago
Views:

1 Derivatives Mat 20 Calculus I D Joyce, Fall 203 Since we ave a good understanding of its, we can develop derivatives very quickly. Recall tat we defined te derivative f x of a function f at x to be te value of te it f fx + fx x. Sometimes te it doesn t exist, and ten we say tat te function is not differentiable at x, but usually te it does exist and, so, te function is differentiable. Derivatives of linear functions. Te grap of a linear function fx ax + b is a straigt line wit slope a. We expect tat te derivative f x sould be te constant slope a, and tat s wat we find it is wen we apply te definition of derivative. f x fx + fx ax + + b ax + b ax + a + b ax b a a a Derivatives of constant functions. We ve actually sown tat te derivative of a constant function fx b is 0. Tat s because a constant function is a special case of a linear function were te coefficient of x is 0. It makes sense tat te derivative of a constant is 0, since te slope of te orizontal straigt line y b is 0. Derivatives of powers of x. We ll directly compute te derivatives of a few powers of x like x 2, x 3, /x, and x. Note tat tese last two are actually powers of x even toug we usually don t write tem tat way. Te reciprocal of x is x raised to te power, tat is, /x x. Also, te square root of x is x raised to te power 2, tat is, x x /2. After computing tese, we ll see tat tey all fit te pattern tat says te derivative of x n is nx n. Tis pattern is called te power rule for derivatives.

Sometimes te it doesn t exist, and ten we say tat te function is not differentiable at x, but usually te it does exist and, so, te function is differentiable. Derivatives of linear functions.

2 First, we ll find te derivative of fx x 2. Tus, te derivative of x 2 is 2x. f x fx + fx x + 2 x 2 x 2 + 2x + 2 x 2 2x + 2 2x + 2x Next, we ll compute te derivative of fx x 3. At one point in te computation, we need to expand te cube of a binomial x + 3, and tat s someting we could do by expanding x + x + x +. Te rule tat gives te expansion of te general power x + n is called te binomial teorem, and tat s related to Pascal s triangle. f x fx + fx x + 3 x 2 x 3 + 3x 2 + 3x x 3 3x 2 + 3x x 2 + 3x + 2 3x 2 Tus, te derivative of x 3 is 3x 2. Note tat we didn t take te it until we cancelled te. Tat s always wat appens. Now, let s try to find te derivative of fx /x. We start te same way as always f x fx + fx /x + /x At tis point, put te numerator over a common denominator, ten simplify te compound 2

Te rule tat gives te expansion of te general power x + n is called te binomial teorem, and tat s related to Pascal s triangle.

3 quotient before continuing. x x+ x+x x x + x + x x + x x + x x 2 Tus, te derivative of /x is /x 2. Tis result fits te power rule mentioned above since we can rewrite it to say te derivative of x is x 2. It s a bit arder to compute te derivative of x since at one point we ave to multiply te numerator and denominator of a quotient by a conjugate, but oterwise it s about te same difficulty. You migt tink tat multiplying and dividing by a conjugate is a trick, but since we do it so often, you sould tink of it as a tecnique. Let fx x. Ten Tus, te derivative of x is of x /2 is 2 x /2. f fx + fx x x + x x + x x + x x + + x x + + x x + + x 2 x x + + x x + + x 2. Again, tis fits te power rule, since it says te derivative x Here s a summary of te computations we ve done so far. fx f x ax + b a x 2 2x x 3 3x 2 /x /x 2 x 2 x 3

It s a bit arder to compute te derivative of x since at one point we ave to multiply te numerator and denominator of a quotient by a conjugate, but oterwise it s about te same difficulty.

4 Most of tese results are special cases of te power rule wic says tat te derivative of x n is nx n. Rules of differentiation. We could go on like tis for eac time we wanted te derivative of a new function, but tere are better ways. Tere are several rules tat apply in broad cases. In fact, we ll eventually find enoug rules for differentiation tat we won t need to go back to te definition in terms of its. So, let s get started. Te sum rule. Let s begin wit te rule for sums of functions. Many functions are sums of simpler functions. For example x 3 +x 2 is te sum of two functions we ve already differentiated. If we can discover te differentiation rule for sums, we ll be able to differentiate x 3 +x 2 witout going back to te definition of derivative. Let f and g be two functions wose derivatives f and g we already know. Can we find te derivative f + g of teir sum f + g? We ll need te definition of derivative to do tat. Wen we apply te definition, we get f + g f + gx + f + gx x. Now, te expression f + gx means fx + gx, terefore, te expression f + gx + means fx + + gx +. We can continue as follows. f + g x fx + + gx + fx + gx fx + + gx + fx gx fx + fx + gx + gx fx + fx + fx + fx gx + gx + f x + g x gx + gx Tus, we ave sown tat te derivative f + g of te sum f + g equals te sum f + g of te derivatives. Tis is a very useful rule. For instance, we can use it to conclude tat te derivative of x 3 + x 2 is 3x 2 + 2x because we already know te derivative of x 3 is 3x 2 and te derivative of x 2 is 2x. Te difference rule. If you just look back troug te previous paragrap and cange some plus signs to minus signs, you ll see tat te derivative f g of te difference f g equals te difference f g of te derivatives. So, for instance, te derivative of x 3 x 2 is 3x 2 2x. Constant multiple rule. On te same level of difficulty as addition and subtraction is multiplcation by constants. If f is a function, and c is a constant, ten cf is te function 4

In fact, we ll eventually find enoug rules for differentiation tat we won t need to go back to te definition in terms of its. So, let s get started. Te sum rule.

5 wose value at x is c fx. We can easily find te derivative of cf in terms of te derivative of f. cf x cfx + cfx c fx + c fx cfx + fx fx + fx c c f x Tus, te derivative of a constant times a function is tat constant times te derivative of te function, tat is, cf c f. Derivatives of polynomials. Wit te elp of te power rule, we can find te derivative of any polynomial. For example, te derivative of 0x 3 7x 2 + 5x 8 is 30x 2 4x + 5. Finding derivatives of polynomials is so easy all you ave to do is write down te answer, but ere are te details so you can see tat we re using all te rules we ave so far. We ll use te abbreviated notation 0x 3 7x 2 + 5x 8 for te derivative of 0x 3 7x 2 + 5x 8. 0x 3 7x 2 + 5x 8 equals, by te sum and difference rules, 0x 3 7x 2 + 5x 8 wic equals, by te constant multiple rule, 0x 3 7x 2 + 5x 8 wic equals, by te power rule and constant rule, 03x 2 72x + 5 0, wic simplifies to 30x 2 4x + 5, te answer. Te product, reciprocal, and quotient rules. Tese tree rules are arder to prove, so we ll put off te proofs for a little bit. First we ll state tem, ten use tem, and finally prove tem. Let f and g be two differentiable functions. Product rule: fg f g + fg Reciprocal rule: g g g 2 f Quotient rule: f g fg g g 2 For our example to illustrate te use of te product rule, let s find te derivative of x x. Here, fx x wile gx x. Ten x x f xgx + fxg x x + x 2 x and tis last expression simplifies to 3 2 x. Wat s important to see in tis example is ow to use te product rule. In words, te product rule says tat te derivative of te product fg of two functions equals te derivative f of te first times te second g plus te first f times te derivative g of te second. 5

Wit te elp of te power rule, we can find te derivative of any polynomial. For example, te derivative of 0x 3 7x 2 + 5x 8 is 30x 2 4x + 5.

6 Unlike sums and differences, te derivative of te product is not te product of te derivatives. Tat is to say, fg does not equal f g. Next, let s ave an example to illustrate te use of te reciprocal rule. Let s find te derivative of 5x 3 x + 2. Tis is te reciprocal of 5x3 x + 2, and we know te derivative of 5x 3 x + 2 is 5x 2. Te reciprocal rule says g g g. 2 In tis example gx 5x 3 x + 2 and g x 5x 2. Terefore, 5x2 5x 3 x + 2 5x 3 x Finally, let s ave an example to illustrate te use of te quotient rule. Let s find te 6x 3 8x 2 derivative of. Te quotient rule says x 2 + 2x + 8 f f g fg. g g 2 In tis example, fx 6x 3 8x 2 and gx x 2 + 2x + 8. We know te derivatives of f and g. Tey are f x 8x 2 6x and g x 2x + 2. Terefore, 6x 3 8x 2 f xgx fxg x x 2 + 2x + 8 gx 2 8x2 6xx 2 + 2x + 8 6x 3 8x 2 2x + 2 x 2 + 2x Of course, we can simplify our answer, but for te purposes of tis example tere is no need to do so. Te derivative of te quotient is complicated. You can remember te rule best in words. It says tat te derivative of a quotient f/g is te derivative f of te numerator times te denominator g plus te numerator f times te derivative g of te denominator, all divided by te square g 2 of te denominator. You ll ave to use te product, reciprocal, and quotient rules several times before you remember tem well. You can actually do witout te reciprocal rule, since it s a special case of te quotient rule, but it comes up often enoug so tat it s wort wile to memorize it. Mat 20 Home Page at ttp://mat.clarku.edu/~djoyce/ma20/ 6

Te reciprocal rule says g g g. 2 In tis example gx 5x 3 x + 2 and g x 5x 2. Terefore, 5x2 5x 3 x + 2 5x 3 x + 2. 2 Finally, let s ave an example to illustrate te use of te quotient rule.

f(a + h) f(a) f (a) = lim

f(a + h) f(a) f (a) = lim Lecture 7 : Derivative AS a Function In te previous section we defined te derivative of a function f at a number a (wen te function f is defined in an open interval containing a) to be f (a) 0 f(a + )