Understanding Derivatives and Their Uses in Calculus
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Learn how limits as x approaches a value are shown graphically, understand derivatives, continuity conditions, and the chain rule in calculus. Explore discontinuous functions and geometric relations between functions and derivatives.
Understanding Derivatives and Their Uses in Calculus
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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