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Chapter Two. Derivatives and Their Uses. Limit as x→3 Shown Graphically. Derivative : 導數. (1) x takes values closer and closer to 3 but never equaling 3 lim ( x →3) (2x+4)=10 lim ( x → c ) f(x) lim ( x → c - ) f(x) 左極限 lim ( x → c + ) f(x) 右極限.
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Chapter Two Derivatives and Their Uses
Derivative : 導數 (1) x takes values closer and closer to 3 but never equaling 3 lim(x→3) (2x+4)=10 lim(x→c) f(x) lim(x→c-) f(x) 左極限 lim(x→c+) f(x) 右極限
(2) the limit of a function may not exist but if the limit does exist , it must be a single number. Define: lim(x→0) (1+x)1/x = L=2.718 lim(x→∞) (1+1/n)n ( lim(x→c) f(x)= L , if and only if ) ( lim(x→c-) f(x)=lim(x→c+) f(x) = L )
求分式極限先化簡(simplifying): • 表示x-value close to 1, not equal to 1 • Ex: =
every polynomial function is continous every rational function is continous , except where the denominator(分母) is 0 Ex : f(x)= f(1)=1-1+1+2-1=2 lim(x→1-) f(x)= lim(x→1-) (x5 –x4 +x3 +2x –1)=2 lim(x→1+) f(x)= lim(x→1+)(x2+x+2 /x+1)=4/2=2 lim(x→1-) f(x)= lim(x→1+) f(x)= f(1) ⇒ f(x) is continous at x=1
slope the average rate of change Instantaneous rate of change
Graphs showing that as Q approaches P, secant becomes the tangent:
1.the line through P and Q call secant line (割線) • 2.let Q 朝P滑動,in the limit as Q→P, The secant line→ the tangent line at P
the average rate of change: Slop(P-Q secant line)= [f(x+h)–f(x)]/ h Instantaneous rate of change : Slop(P-tangent line)= lim(h→0) [f(x+h)–f(x)]/ h The Derivative : f’(x) : the slope of f at x;measure how steeply the curve rises at x ;the Instantaneous(瞬間的) rate of change of f at x f’(x) =lim(h→0) [f(x+h)–f(x)]/ h
tangent line(切線) to the curve at P(1,1) the tangent line fits the curve so closely , it called the best liner approximation to the curve
A. c is constant d/dx c =0 f’(x)=lim(h→0)h(x+h)-f(x)/h =lim(h→0) c-c/0=0 B. d/dx c‧f(x)=c‧f ’(x) 令g(x)= c‧f(x) g’(x)= lim(h→0)[g(x+h)-g(x)] /h = lim(h→0)[c‧f(x+h)- c‧f(x)] /h = lim(h→0)c‧[f(x+h)- f(x)] /h = c‧lim(h→0)[f(x+h)- f(x)] /h = c‧f(x)
C. d/dx [f(x)±g(x)]=f’(x) ± g’(x) 令p(x)=f(x)±g(x) p’(x)= lim(h→0) p(x+h)-p(x)/h =lim(h→0) [f(x+h)±g(x+h)] - [f(x)±g(x)] /h = lim(h→0) [f(x+h)-f(x)] ± [g(x+h)-g(x)] /h = lim(h→0) f(x+h)-f(x)/h ±lim(h→0) g(x+h)-g(x)/h = f’(x) ± g’(x)
D. d/dx [f(x)g(x)]=f’(x)g(x)+f(x)g’(x) 令p(x)=f(x)g(x) p’(x)= lim(h→0) [p(x+h)-p(x)]/h = lim(h→0) [f(x+h)g(x+h)-f(x)g(x)+f(x)g(x+h)- f(x)g(x+h)] /h =lim(h→0) g(x+h)[f(x+h)-f(x)]+f(x)[g(x+h)-g(x)] /h =lim(h→0) f(x+h)-f(x)/h‧lim(h→0) g(x+h)+ lim(h→0) g(x+h)-g(x)/h‧f(x) =f’(x)g(x)+f(x)g’(x)
E. If f’ g’ exist with g(x)≠0 , g(x)=f(x)/g(x) , then g’(x)=f’(x)g(x)-f(x)g’(x)/[g(x)]2 g’(x)= lim(h→0) g(x+h)-g(x)/h = lim(h→0) 1/h‧[f(x+h)/g(x+h)-f(x)/g(x)] = lim(h→0) 1/h‧[f(x+h)g(x)-f(x)/g(h+x)+f(x)g(x)-f(x)g(x)] /g(x+h)‧g(x) = lim(h→0) 1/g(x+h)g(x)‧[ g(x)[f(x+h)-f(x)] /h-f(x)[g(x+h)-g(x)] /h ] =1/[g(x)]2 [f’(x)g(x)-f(x)g’(x)] =f’(x)g(x)-f(x)g’(x)/[g(x)]2
F. d/dx xn =n.xn-1 (1) n:正整數 (positive integer) (x+h)2 =x2 +2xh+h2 (x+h)3 =x3 +3x3h+3xh3+h3 (x+h)n =xn +nxn-1 h+h2 [1/2(n-1)h x2 +h2 ….] =xn +nxn-1 h+h2.p f’(x) = lim(h→0) (x+h)n – xn /h = lim(h→0) nxn-1 h+h2 p /h = lim(h→0) nxn-1 +h.p = nxn-1
(2) n:負整數 (negative integer) let n=-p (positive integer) d/dx (xn)=d/dx (1/xp) =-p.xp-1.1 /x2p =-p. xp-1-2p =-p.x-p-1 = nxn-1
The Chain Rule <合成函數微分> d/dx f(g(x))=f’(g(x)).g’(x) let y=f(u) , u=g(x) ⇒ y=f’(g(x)) dy/du =f ’(u) , du/dx =g’(x) dy/dx=d/dx f(g(x)) ∥ dy/du.du/dx =f’(x).g’(x) =f’(g(x)).g’(x)
Ex: f(u)=u2 , u=g(x)=4-x f(g(x))=(4-x)2 =16+ x2 -8x d/dx f(g(x)) =2(4-x)(4-x)’ =(8-2x)(-1) = 2x-8
Non differentiable Functions 不可微函數 function that can’t be differentiated at certain values Ex: f(x)=|x|, at x=0 lim(h→0) [f(0+h)-f(0)] /h = lim(h→0) [f(h)-0] /h= lim(h→0)|h|/ h ① lim(h→0+) |h|/ h = h/h=1 ② lim(h→0-) |h|/ h = -h/h=-1 lim(h→0+) = lim(h→0-) f(x) ∴ f’(0) doesn’t exist
※ f will not be differentiable at c f has a cornerpoint at x = c f has a vertical tangent at x = c f is discontinuous at x = c
※ 可微性與連續性 A ⇒ B ①f’(c)存在⇒ f 在 x=c 連續 lim(x→c) f(x)-f(c)/x-c存在 , f’(c) lim(h→0) f(c+h)-f(c)/n x-c=h→0 ⇒c+h=x x-c→0 x→c <證明> lim(x→c) f(x)=f(c) lim(x→c) f(x)- f(c)= lim(x→c) f(x)-f(c)/x-c.(x-c) =f’(c) lim(x→c) (x-c) =f’(c).0 =0 ⇒ lim(x→c) f(x)=f(c) , ∴ f在x=c 連續
② ⇐ x B ⇐ A (x) ex: f(x)=|x| at x=0 f is continous , at x=0 f’(0) doesn’t exist
Venn Diagram Showing Relation Between Continuous and Differentiable Functions