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# Rutgers - Calculus 640 - Class Notes - Week 5

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Rutgers - Calculus 640 - Class Notes - Week 5

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##### Description: These notes cover from Section 3.3 to 3.9 excluding 3.4
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Unformatted text preview: 02419124 Section 3.3 differentiable, then f is Product Rule: ff, 9 ore differenliable and 10(x)(-)) = f'(*)g(x) + f(x) g'(x) Exi- find. IO3 1 " (~) = (4x})(cy) + (x4) (e) f'(x) g(x) f(x) g'(-) f) 9(-) 6) 56) = (2x2-7)(x+40%) 7 5)= {22)-o](5x +45) EC) 90) c)f(x) = *{(), f(43):3,"l2)-7 Find (2) Pec) glx) Let 90) be equal to to? (x) = f'(x) + f(x) (2x) games goed F'(@) = f'(2) (22) + f(2) (262) = (-+)(4) + (3)(4) -16 Quotient Bule:- r_f99. are differentiable, then f is differentiable andi d (162) = f(x) g(x) - f(x) '() dx 907 (x)}? One way to remeber it is lodhi - hidlo (10) 10 = bottom hi top de derivative6: find the derivative a) r() - 2_ ) 4x5469 (2) 297 96) 10-) g'() F'@) = (2x)(42516) - (2) [4(534) (4+5+c)? gonja b) f(x) = x - 2* .fl.) 3ex + = 9x) (c) 916) r() - G -23ext) - F-22) (3*** (3*3 458) 96) Example h(x) = xfG) Find 1 (3) if f(3)--| f(3) = 4, g(3)-5, and g'(3) = -2 h'() =*(*F(-) (96-)) ( f(-) 1: (96-)) (g(x)) - (1) f()+ xf?@]g(x) = x f(x) g'(x) [g(x)]? 1113)= [f(3) + 3 f'13)] g(3) - (3) f(3)g'(3)! ( 9(3))?h (3) = [(-1) +3(4)]5) - 3 (-1)(-2) (5) find the longent line toh when x=3 slope (3) - 4 (3) - 3.468) - 360- Point 13,-3) tengont_line; y+ 3 h (x-3) - Postponed until after Midterm 3.4 is Section 3.5 f() is differentiable. I t Secend derivative of fis p" (x) = + (f*(x)) differentiable, then the Similarlys fos () -d (F"(x)) pens (*), fos) (*) , -.. Exomple :- find f** () if f(- f (x) - 3-4 F(x)-4(3)x5 -12x-5 f"(x) = -5(-12) x6 = 60x6 f ) = -6 (60) x-7 ---- -360x-7

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