f DERIVATIVES NOTES HARRIS MATH CAMP 208. Introduction Reding: Section 2. The derivtive of function t point is the slope of the tngent line to the function t tht point. Wht does this men, nd how do we clculte it? We pproimte the tngent line with secnt lines, nd use limits! " " I 5 A = Ath Slope = rise = fi ) C ) Definition. The derivtive of function f t is f 0 f() () = lim! f() f( + h) h! h f(). when tht limit eists. We cn lso define the righthnded nd lefthnded derivtives t by tking the righthnded nd lefthnded limits, respectively. For this limit to eist, the function f hs to be continuous t. We cn see why tht is true for the cse of jump discontinuity by noting tht the righthnded limit pproches + nd the lefthnded limit pproches, so the limit does not eist.
2 HARRIS MATH CAMP 208 Definition. A function is di erentible t if f 0 () eists. Emple. () The function f() =c is di erentible everywhere, nd f 0 () =0. By the definition, f 0 f() () = lim! f() c! c 0!! 0=0. (2) The function g() = is di erentible everywhere, nd g 0 () =. From the definition, g 0 g() () = lim! g()!! =. (3) The function h() = 2 is di erentible everywhere, nd h 0 () =2. From the definition, h 0 h() () = lim! h() 2 2!! ( )( + ) + = + =2.! Note tht in this cse, the derivtive of h depends on our choice of. In other words, the slope of the tngent line chnges s we chnge our vlue. Notice tht it is not true tht every continuous function is di erentible! Consider the function f() =, or f() = 3p. Emple. The function f() = p is di erentible everywhere ecept = 0, nd f 0 () = 2 p. So the slope of the tngent line t =is 2, nd the eqution of the tngent line t =isy = 2 ( ), or y = 2 + 2. Theorem.. (Power Rule) The function f() = n hs derivtive f 0 () =n n, nd is di erentible for >0. 2. Derivtive Rules Reding: Sections 2.22.4, Chpter 4 Theorem 2.. Suppose tht f() nd g() re di erentible t, ndk is rel number. () (Linerity) (f + kg) 0 () =f 0 ()+k g 0 () (2) (Product Rule) (f g) 0 () =f()g 0 ()+f 0 ()g() (3) (Quotient Rule) ( f g )0 () = g()f 0 () g 0 ()f() g() 2 (4) (Chin Rule) (f g) 0 () =f 0 (g()) g 0 () A corollry of this is tht polynomils re di erentible everywhere, nd rtionl functions re di erentible everywhere on their domin.
DERIVATIVES NOTES 3 Emple. The function f() =( 2 + ) 3, then f 0 () = 3( 2 + ) 2 +2. Emple. Suppose f() =. Then f 0 () = lim h!0 +h h h!0 ( h ) h = lim h!0 h h = f 0 (0). So if we wnt to find the derivtive of f() =, we need to know the derivtive t = 0. We define the e number e s the unique number such tht = lim h h!0 h. Thus if g() =e,theng 0 () =g() =e. More generlly, f 0 () = ln, where ln = log e. There re other wys to define the number e: () e n! + n n (2) e = P n=0 n! (3) e =2.7828... Emple. Suppose f() = log (). Then we cn find f 0 () by using the fct tht eponentils nd logrithms re inverse functions. By the chin rule, if f g() =, then f 0 () = g 0 (f()).so Reding: Section 2.3 f 0 () = log() ln = ln 3. HigherOrder Derivtives Suppose tht function s(t) gives the loction of n object t time t. Then the derivtive s 0 (t) gives you the chnge in loction over time. In other words, s 0 (t) =v(t) isthevelocity of the object. Furthermore, the derivtive of the velocity, v 0 (t) =s 00 (t), tells how the velocity chnges, or the ccelertion of the object. If we wnt to clculte the ccelertion of n object, then we need to find the second derivtive of the loction function.
4 HARRIS MATH CAMP 208 Emple. Suppose tht n object is moving bck nd forth on number line, ccording to the rule tht s(t) =t 2 ln t. Then the velocity t time t is given by v(t) =s 0 (t) =2t t. To find the ccelertion, we tke the second derivtive of s(t), which is the sme s the derivtive of v(t) : (t) =v 0 (t) =s 00 (t) =2+ t 2. 4. Optimiztion Reding: Sections 3.3.2 The derivtive of function gives you informtion on the growth of function. If the derivtive is positive, the function is incresing. If the derivtive is negtive, the function is decresing. Definition. A function f is incresing on n intervl (, b) if: for every nd y in (, b) with<y,we hve f() <f(y). The function is decresing on n intervl (, b) if: for every nd y in (, b) with<y, we hve f(y) <f(). Incresing Decresing LET y By looking t the intervls on which function is incresing or decresing, we cn find the mimum nd minimum vlues chieved by the function. y Definition. A point c is locl mimum of function f if f(c) f() for ll in some intervl (, b) contining c. Similrly, point d is locl minimum of function f if f(d) pple f() for ll in some intervl (, b) contining d. We cll mim nd minim the etrem of function. Notice tht the locl etrem of function cn only occur when f switches between incresing nd decresing, or t n endpoint of the rnge. In other words, locl etremum cn only occur if f 0 chnges sign, or t n endpoint of the rnge of f. Therefore, locl etremum cn only occur t if f 0 () = 0 or is undefined, or if is n endpoint of the rnge of f.
DERIVATIVES NOTES 5 Definition. The criticl points of function f re the vlues of t which f 0 () = 0 or f 0 () isundefined. Theorem 4.. (First Derivtive Test) Suppose tht c is criticl point of f. Then c is locl mimum if f 0 () > 0 for <c,ndf 0 () < 0 for >c, locl minimum if f 0 () < 0 for <cnd f 0 () > 0 for <c,nd not locl etremum if f 0 () hs the sme sign on either side of c. Definition. A point c is globl mimum of function f if f(c) f() for ll other in the rnge of f. Similrly, point c is globl minimum of function f if f(c) pple f() for ll in the rnge of f. Notice tht every globl etremum is lso locl etremum. Emple. Suppose f() = 2.Thenf 0 () =2, so f is decresing on the intervl (, 0), nd incresing on the intervl (0, +). Thus f hs (globl) minimum t = 0. But it doesn t hve globl mimum. " Afl ) " '. t fi ) Similrly, the second derivtive cn give you informtion bout the grph of function. Definition. A function f is concve up on n intervl (, b) iff 00 () > 0 on tht intervl. It is concve down if f 00 () < 0 on the intervl. A point c t which the concvity chnges is clled point of inflection. As before, points of inflection cn only occur if f 00 (c) = 0 or f 00 (c) is undefined. Note tht c cn only be point of inflection if f(c) is defined nd f is continuous t c. Theorem 4.2. (Second Derivtive Test) Suppose f is di erentible function nd f 0 (c) =0. Then: f hs reltive mimum t c if f 00 (c) < 0, nd f hs reltive minimum t c if f 00 (c) > 0. If f 00 (c) =0, then the test is inconclusive.
6 HARRIS MATH CAMP 208 Reding: Section 2.6 5. Implicit Differentition nd Relted Rtes We hve been using the nottion f 0 () to denote the derivtive of f t. But there re more wys to denote this. We cn lso sy f 0 () = dy d, nd t specific point = we sy f 0 () = dy d =. The first nottion is useful when we hve function written eplicitly, s in f() = 2 + e. But we cn lso write the function implicitly, such s 2 y 3 + y = e. Notice tht you hve to be creful bout domin nd rnge of your function when it is written implicitly. In fct, you my not hve function t ll! For emple, y 2 = is not function with input nd output y, since for = there re two yvlues which stisfy the eqution. We cn resolve this by, for emple, lwys choosing the positive yvlue. However, you must be creful! Then we cn di erentite n implicit function using implicit di erentition. Emple. Let 2 y 3 + y = e. Compute dy d =0. Then: d d (2 y 3 + y) = d d (e ) d d (2 y 3 )+ d (y) =e d ( d d 2 )y 3 + 2 ( d d y3 )+y 0 = e 2y 3 +3 2 y 2 y 0 + y 0 = e. Now t = 0, we hve 0 y 3 + y = e 0 =, so y =. Then we hve e 0 ==2 0 () 3 +3 0 2 2 y 0 + y 0 = y 0, nd so dy d =0 =. This llows us to solve problems where we wnt to compre the rte of chnge of two relted functions. Emple. Suppose tht we hve cylinder with rdius 2in tht is full of wter, nd we punch hole in the bottom so tht the wter flows out, t rte of cubic inch per second. At wht rte is the height of wter in the cylinder chnging?
DERIVATIVES NOTES 7 Solution. The mount of wter in the cylinder is given by V (t) =B h(t) =( 2 2 )h(t). Since the wter is flowing out t rte of cubic inch per second, then V 0 (t) = for ll vlues of t. By implicit di erentition, V 0 (t) =4 h 0 (t), nd thus h 0 (t) = 4, nd the height of wter in the cylinder is decresing t rte of 4 second. cubic inches per Another use of implicit di erentition is to simplify equtions tht would otherwise be complicted to di erentite. Emple. Suppose y =( 2 ). To di erentite this, tke the nturl logrithm of both sides: ln y =ln( 2 ) = ln( 2 ). Now use implicit di erentition: Simplifying, we get y y0 =ln( 2 ) + 2 2. y 0 = y(ln( 2 ) + 22 2 )=(2 ) ln( 2 ) + 22 2 Emple. Let y = 2 5 + 4 + 3 + 2 ++. Using the Quotient Rule on this would be nnoying. Insted, multiply both sides of the eqution by the denomintor: y( 5 + 4 + 3 + 2 + + ) = 2. Using implicit di erentition, we get y 0 ( 5 + 4 + 3 + 2 + + ) + y(5 4 +4 3 +3 2 +2 + ) = 2, nd fter simplying we see y 0 = 2 y(54 +4 3 +3 2 +2 + ) 5 + 4 + 3 + 2. + +