Eam Rviw Solutions. Comput th following rivativs using th iffrntiation ruls: a.) cot cot cot csc cot cos 5 cos 5 cos 5 cos 5 sin 5 5 b.) c.) sin( ) sin( ) y sin( ) ln( y) ln( ) ln( y) sin( ) ln( ) y y y cos( ) ln( ) sin( ) y y sin( ) ln( ) sin( ) ln( ) sin( ) cos( ) ln( ) sin( ).) sin cos cos ( ).) y tan( y) y tan( y) y tan( y) y y y y y y y y y tan( y) y sc ( y) t t ln() t ln() 0 t f.) t
Eam Rviw Solutions g.) 0 5 sc6 cos 0 5 0 0 sc(6 ) tan(6 ) (6) ( ) 0 h.) cos y y y y sin y y ( y ) y y y y sin y ( y) siny y y y ( y ) sin y y y sin y y ( y ) sin y y y sin y y y siny ( y) sin y y. Fin th following: a. Th quation of th tangnt lin to th curv y 5 ( ) at th point whr. Th slop of th tangnt lin is th rivativ valuat at =. 5 y () y 5( ) () y () 5() () 0 So th quation for th tangnt lin is: y 0( )
Eam Rviw Solutions b. Th quations of th lins tangnt to y an prpnicular to th lin through th points (6, 6) an (, 8). (Hint: If a lin has a slop of m, a lin prpnicular to th first lin has a slop of ). m Th lin through (6,6) an (, 8) has a slop of quations of all tangnt lins with a slop of y ( ) y ( ) ( ) ( ) ( ) 0 or 86 6 6. So w want th tangnt lins at (0,0) an (, ), which ar: y 0 ( 0) an y ( ) so w want th c. Fin th quation of th tangnt lin to th curv y y 7 at th point (, ). y y 7 y y ( y ) y 0 y y y y y y y y (,) () 5 So th quation for th tangnt lin is: y ( ) 5
Eam Rviw Solutions 5. Fin th points on th curv y whr th tangnt lins ar horizontal an vrtical. y y 6y 0 y 6y So th tangnt lin is horizontal whn 0 y y 8 8 y y 8,, 8 8 An th tangnt lin is vrtical whn 6y 0 y 0 0 0,0,0
Eam Rviw Solutions 6 5. A particl is moving along a curv y. a. As th particl passs through th point,, what is th vlocity of th particl? y y () y() () b. As th particl passs through th point,, what is th acclration of th particl? y y () ()() 9 9 9 9 y() () (8) c. Whn is th particl stationary? y 0 if 0 which nvr happns, so th particl is nvr stationary 6. Lt H f f, an G H. Also lt f 5, f, f '5 7. f ' 0, an a. Fin H' ( f( f( )) f( f( )) f( ). H' 5 f( f(5)) f(5) f() 7 07 70 b. Fin G' ( H( )) ( H( )) ( H( )) G' 5 ( H(5)) ( H(5)) f( f(5)) 70 f() 70 70 0
Eam Rviw Solutions 7. Fin a an b so that th function f iffrntiabl. For f() to b continuous, w must hav lim f( ) lim f( ) lim 6 lim ab 6, is both continuous an a b, 6() () a() b 8 ab For f() to b iffrntiabl, w must hav f( h) f() f( h) f() lim lim h0 h h0 h 6( h) ( h) a() b a( h) b a() b lim lim h0 h h0 h But 6() () a() b bcaus w know th function is going to b continuous at. Making this substitution, w gt: 6( ) ( ) () ( ) () lim h h a b a h b a b lim h0 h h0 h But this is just th finition of th rivativ for ach function, so w gt: 6 a b a () a a Substituting back into th first quation givs 8 ab 8 () b 8 b b 7
Eam Rviw Solutions 8 8. Fin th following limits. Your answr shoul b a numbr,, or DNE. a. sin sin sin sin lim lim lim lim lim 0 tan 7 0 tan 7 0 tan 7 0 0 tan 7 cos(7 ) lim lim lim lim 0 0 sin 7 tan 7 lim cos(7 ) 0 0 cos7 sin(7 ) sin(7 ) 0 7 7 lim lim lim lim lim 0sin(7 ) 0 sin(7 ) 7 0sin(7 ) 0 7 0 7 7 b. sincos sin cos sin cos lim lim lim lim 0 0 0 0 0 0 9. Lt f ( ) b fin by th graph on th right. f ( ) Lt g ( ) 5. Lt h ( ) g ( ) a) Fin f ( g ( )) whn. f( g( )) f( g( )) g( ) g so g ( ) 5 () 5() () g so g ( ) 5 () 5() Thus, w hav f( g()) g() f() () () 9 b) Fin h ( ) whn. f( ) f( ) g( ) f( ) g( ) h ( ) g( ) g ( ) g so g ( ) 5 () 5() () 6 g( ) 5 so g() 5() 58 f() g() f() g() ( ) (6) () (58) 08.075 g() 6
Eam Rviw Solutions 9 ( ) c) Fin h whn. f( ) f( ) g( ) f( ) g( ) h ( ) g( ) g ( ) f ( ) g( ) f( ) g ( ) h ( ) g ( ) g ( ) f( ) g( ) f( ) g( ) f( ) g( ) f( ) g( ) g( ) f( ) g( ) f( ) g( ) g( ) g( ) g ( ) 5 so g() 5() () 6 g so g g( ) 0 so g() 0() 60 ( ) 5 () 5() 58 f(), f(), f() 0 h ( ) f() g() f() g() f() g() f() g() g() f() g() f() g() g() (0) (6) ( )(58) ( ) (58) () (60) 6 ( ) (6) () (58) 6 (58) 6 7 7 609608 5876 67966 60966676 77760 696 770976.5909 67966 67966 67966 g() g()
Eam Rviw Solutions 0 y 0. Comput th following rivativs using logarithmic iffrntiation: 0 cos ln 0 cos ln ln( y) ln 0 cos ln 0 0 ln( y) ln cos ln ln ln( y) ln ln cos ln ln ln( y) 0ln lncos ln ln ln( y) 0 ln ln cos ln y 0 sin y cos ln y 0 cos ln 0 sin cos ln
Eam Rviw Solutions Quick Summary of th Matrial W strongly suggst you mak your own stuy gui using your book an class nots. Hr ar a fw things you may want to inclu:. Anything from Eam that your ar still unsur of an sms to hav com up in class, homwork, or wbwork sinc am.. Th Dfinition of th Drivativ Th rivativ is th instantanous rat of chang of a function Thr ar two possibl typs of answrs for a rivativ o If w ar ask th rivativ at a spcific point, our answr will b a numbr. o If w ar ask for th rivativ without bing givn a spcific point, our answr will b a function. Thr ar two main forms for th finition of th rivativ. Thy ar quivalnt. f ( ) f ( t) o f '( t) lim t t f ( h) f ( ) o f '( ) lim h0 h How to o th rivativ bas on th finition for th following functions: o Polynomials ampl f ( ) o Roots ampls f ( ) (multiply by th conjugat) o Rational Functions ampls f ) ( or. Diffrntiability of a Function A function is iffrntiabl at a point a if f ( ) f ( ) f ( a) lim L a f ( ) f ( a) a lim a a f ( ) f ( a) lim M a a an L M If a function is iffrntiabl at a thn th function is continuous at a If a function is iscontinuous at a thn it is not iffrntiabl at a A continuous function n not b iffrntiabl at a o a is a cusp or cornr point th limit will fail bcaus L M o th limit at a will prouc an unfin ROC, i a vrtical tangnt. Tabl of Ky Drivativs: Eponnt an Log functions a a ln a ln Trigonomtric functions sin cos cos sin tan sc csc csc cot sc sc tan cot csc Invrs Trig functions sin cos tan
Eam Rviw Solutions 5. Spcial Limits sin( ) lim lim 0 0 sin( ) cos( ) lim 0 0 lim lim 0 6. Drivativ Ruls c 0 rivativ of ANY constant ( c f) c f rivativ of a constant tims a function n ( ) n n th Powr Rul ( f g) f g sum or iffrnc of functions f g fg f g th Prouct Rul th Quotint Rul f fgfg g g [ f ( g)] f( g) g th Chain Rul 7. Implicit Diffrntiation Us whn it is impossibl or impractical to solv a function in two variabls to b in trms of on variabl. Eampl: y. In this ampl, w woul say that y is implicitly fin as a variabl or function of. Diffrntiat both an y as you woul alon, but whn you iffrntiat y, apply th y chain rul to gt point y or, on th curv. y ' (ithr is fin). Thn solv for y. To valuat th rivativ w woul n a E. y y y 0 y y y y y 8. Logarithmic Diffrntiation Us whn th function is complicat or for functions with an in th ponnt. Tak th log of both sis, simplify with log proprtis, iffrntiat (implicit chain rul on y will always happn on th lft si), thn solv for y '.