函數的微分

函數的微分
1-3 函數微分式

(甲)基本函數的微分公式

證 :
(1) Let f(x) = xⁿ   ,   f'(x) = lim f(x+h) - f(x) / h = lim [ (x+h)ⁿ  -  xⁿ  ] / h  = lim  ∑ [ _nC_k  x^n-kh^k ] - xⁿ  / h  =  lim  [ _nC_n-n  x^n-nhⁿ  + _nC_n-1  x^n-n+1h^n-1  + _nC_n-2  x^n-n+2h^n-2  +  ... + _nC₁  x^n-1h¹ + _nC₀  xⁿh⁰  - xⁿ  ] / h  = lim h [_nC_n-n  x^n-nh^n-1  + _nC_n-1  x^n-n+1h^n-2  + _nC_n-2  x^n-n+2h^n-3  +  ... + _nC₁  x^n-1h⁰   ] / h  = lim [_nC_n-n  x^n-nh^n-1  + _nC_n-1  x^n-n+1h^n-2  + _nC_n-2  x^n-n+2h^n-3  +  ... + _nC₁  x^n-1h⁰   ] = 0 + 0 + ...+ 0 + _nC₁  x^n-1h⁰ = _nC₁  x^n-1 = n x<sup>n-1

另一法是用數學歸納法. 

(2) 令 f(x) =</sup> x^1/n  , f'(x) = lim [ f(x+h) - f(x)] / h = lim [(x+h)^1/n - x^1/n ] / h  = lim [ (x+h) - x ] / h [ (x+h)^n-1/n + (x+h)^n-1/nx^1/n +(x+h)^n-2/n x^2/n+(x+h)^n-3/n x^{3/n  ....}+ x^n-1/n] = lim h/h[ (x+h)^n-1/n + (x+h)^n-1/nx^1/n +(x+h)^n-2/n x^2/n+(x+h)^n-3/n x^{3/n  ....}+ x^n-1/n] = lim 1/ [ (x+h)^n-1/n + (x+h)^n-2/nx^1/n +(x+h)^n-3/n x^2/n+(x+h)^n-3/n x^{3/n  ....}+ x^n-1/n] =  1/ nx^n-1/n = 1/nx<sup>1-1/n       


(3) f(x) = c , c is a constant  then f'(x) = 0 ( directive of any constant is 0 ) 

(4)(5) prove them via formulas . 


(乙)導數的四則運算</sup>

證: 

令 h(x) = f(x)g(x) , h'(x) = lim [ h(x+k) - h(x) ]/ k  = lim [ f(x+k)g(x+k) - h(x) ] / k  = lim f(x+k)g(x+k)  - f(x+k)g(x) + f(x+k)g(x)  - f(x)g(x) / k  = lim f(x+k)[g(x+k) - g(x)] + [f(x+k)-f(x)]g(x) / k = f(x)g'(x) + f'(x)g(x) 

According to the rule , we can calculate it .

(b) 可以使用數學歸納法證之. 

(c) 套用(b)的結果

h(x) = g(x)⁵       g(x) = x⁵ + 2x + 3   , h'(x) = d/dx [g(x)⁵] = 5g(x)⁴g'(x) ,  g'(x) =  5x⁴ + 2 

       

Let f(x) = x^r , According to the definition of directive , f'(x) = lim f(x+h) - f(x) / h = lim [ (x+h)^r - x^r ]/ h  . By binomial identity  (x+h)^r  = ∑ _rC_k x ^kh ^r-k  , 
lim [ (x+h)^r - x^r ]/ h  = lim [ ∑ _rC_k x ^kh ^r-k  - x ^r ] / h = lim [_rC_r x ^rh ^r-r  + _rC_r-1 x ^r-1h ^r-r+1 + ... + _rC₁ x ¹h ^r-1 + _rC₀ x ⁰h ^r - x ^r ] / h = lim h [ _rC_r-1 x ^r-1h ^r-r+1-1 + ... + _rC₁ x ¹h ^r-1-1 + _rC₀ x ⁰h ^r-1 ] / h = lim [ _rC_r-1 x ^r-1h ^r-r+1-1 + ... + _rC₁ x ¹h ^r-1-1 + _rC₀ x ⁰h ^r-1 ] = lim [ _rC_r-1 x ^r-1h ^r-r+1-1 + ... + _rC₁ x ¹h ^r-1-1 + _rC₀ x ⁰h ^r-1 ] = r x ^{r-1       
Q.E.D.} 

證:  

f(x)/g(x) can read as f of x over g of x  .   
f(x)/g(x) = f(x) × g(x)^-1   

Let f(x)/g(x) = h(x) 

^{According to  above the rule of directive ,  h(x)'  = [} f(x) × g(x)^{-1 ] '  =}  f'(x) × g(x)^-1  + (g(x)^-1)^'×f(x) =   f '(x) × g(x)^-1 + (-1)g(x)^-2×f(x)×g'(x) =  
f '(x) × g(x) - f(x)×g'(x) / g(x)<sup>2   

So d/dx  (h(x)) | x = a   = [</sup> f '(a) × g(a) - f(a)×g'(a) ] / g(a)²    

  

Let h(x) = f(x)/g(x) , h'(x) = [f '(x) × g(x) - f(x)×g'(x) ] / g(x)^{2   
f'(x) = 2x , g'(x) = 2x +1}   

Let  f(x) = (x^2 + x + 1)^-1   

ln f(x) = - ln(x^2 + x + 1) 

f'(x)/f(x) = - d/dx ( x^2 + x + 1 ) / ( x^2 + x + 1)  

f'(x) = [ f(x) / ( x^2 + x + 1) ] × (2x+1) = f(x)² × (2x+1) 

  

^{(1)(2) [f(x)g(x)]' = f(x)g(x)'  + f(x)'g(x) 
(3)  [} f(x)g(x)h(x)] ' = f '(x)g(x)h(x) + f(x)[g(x)h(x)] '  =    f '(x)g(x)h(x) +  f(x)[g'(x)h(x) + g(x)h'(x) ]  =  f '(x)g(x)h(x) +   f(x)g'(x)h(x) + f(x)g(x)h'(x) 

Suppose that there exist n functions , f1,f2,f3,f4, ... , fn , then (f1× f2× f3× f4 ...  × fn ) '  = f1' × (f2× f3× f4 ...  × fn) +  f2' × (f1× f3× f4 ...  × fn) + ... +  fn' × (f1× f2× f3 ...  × fn-1) = ∑ (fi)' Π fj , i = 1,2,3, ... , n and j = 1,2,3, ... ,n - {i} . 

tanx = sinx / cosx , 
Let f(x) = sinx , g(x) = cosx , h(x) = f(x)/g(x)  
(tanx)' = h'(x) = g(x)f'(x) - f(x)g'(x) / (g(x))^2  
(secx)' = (1/cosx)' =  - d/dx(cosx)/ (cosx)^2  = sinx /(cosx)^2 = (sinx/cosx) (1/cosx) = tanx secx 

(1) df/dx 
(2) using the rule [f(x)g(x)]'
(3) same above 
(4)  (f(x)^n)' = (nf(x)^n-1)f'(x) 

cot x = 1/tanx = cosx/sinx , (cotx)' = (cosx/sinx)'   
(cscx)'  = (1/sinx)' = d/dx(sinx)/ (sinx)^2  


(丙)連鎖法則

  Using chain rule . 

    f(x) = sin²x  = (sinx)²  
     g(x) = sinx , f'(x) = 2sinx d/dx(sinx) = 2sinx(cosx) = 2sinxcosx 

(f(X))ⁿ  = f(x)f(x)... f(x) n times multiplication
d/dx((f(X))ⁿ) = d/dx ( f(x)f(x)... f(x) ) = f'(x)f(x)... f(x)+f'(x)f(x)... f(x)+f'(x)f(x)... f(x)+ ... +f'(x)f(x)... f(x) = n(f(X))^n-1f'(x) . 

But we can use chain rule to evaluate it ? Basically , using chain rule to evaluate the directive of function. The function must be  a composition of functions , that is f∘g form . According to the question , can we use chain rule  ? 

f(x) = ⁵√x  , g(x) = x ⁴ + 3x ²- x + 5 ; f∘g(x) = ⁵√x ⁴+ 3x ²- x + 5  

(f∘g(x))'  = f '(g(x))g'(x) = (4x ³ + 6x - 1) ⁵√(x ⁴+ 3x ²- x + 5) ^-4

  

[(f(x))ⁿ]' = n[(f(x))^n-1]f(x)^'     

(1) f(x)g(x)
(2) (f(x))^3 
(3) f∘g(x) 
(4) f(x)g(x) 
(5) f∘g(x) 

(1) h(x) [f∘g(x)] = h(x)f(g(x)) , [h(x)f(g(x))] '  = h(x)f'(g(x))g'(x) + h'(x)f(g(x)) 

(2) [f(x)/g(x)]'  = g(x)f'(x) - f(x)g'(x) / [g(x)]^2 
(3) 同上 

f(x) is differentiable . 
[f(x-1/x+1)]' = f'(x-1/x+1) d/dx(x-1/x+1) = f'(x-1/x+1) [(x+1) - (x-1)]/ (x+1)^2  = f'(x-1/x+1) 2/(x+1)^2 = 1 
 f'(x-1/x+1) = 1/2 (x+1)^2  , 
 f'(x-1/x+1) | x = 1   =  f'(0) = 1/2 (1+1)^2 = 2 

Chain Rule for composition function 

f(g(X))' = f'(g(x))g'(x)
 f'(g(x)) = 1/4(x/x+1)^(-3/4) , g'(x) = d/dx(x/x+1) = [ (x+1) - x ]/ (x+1)^2 = 1/(x+1)^2.

f(x) = (x/3x+1)^(1/4) , f'(x) = 1/4 [(x/3x+1)^(-3/4)] [d/dx(x/3x+1)]  

Using Chain rule and let h(x) = f(g(x)) , g(x) = 2x + √ x ²+1

Using Chain rule and f'(x) | x = 3

Using [(f(x))ⁿ]' = n[(f(x))^n-1]f(x)^'  

假設該切線方程式為 y = mx + b , m =2 . 而其切點為(c,d) 
故即求 f(x) 在  x = c 上的斜率 m . y' = x² -  x , m = c² -  c 
2 =  c² -  c 
 c² -  c - 2 = 0 
(c- 2)(c+1) = 0 
c = 2 或 c= -1 代回 f(x) , d = -1/3 - 1/2 + 1/3 = -1/2 or d = 8/3 - 2 + 1/3  = 1 
(2,1) 或 (-1,-1/2) 為切點 
(2,1) 代入 y = 2x + b , b = -3 , y = 2x-3
(-1,-1/2) 代入  -1/2 = -2 + b , b = 3/2 , y = 2x+ 3/2 , 2y = 4x +3  

首先, 必須對曲線f(x) 求斜率, f'(x) = x^2 + 2x . 題目要求斜率為最小的切線,如何得知斜率為最小 ? 

  



先求 y' , 再求 y' | x = 1 上的切點斜率 , y'(1) .而-1/y'(1) 為法線斜率.  

f(x) = (x^2 - 1) / (x^2+x+1) , f'(x) = (x^2+x+1)(2x) - (x^2 - 1)(2x+1) / (x^2+x+1)^2  = 2x^3 + 2x^2 + 2x - (2x^3 - 2x + x^2 -1 ) /(x^2+x+1)^2 =   2x^3 + 2x^2 + 2x - 2x^3 + 2x -  x^2 +1 / (x^2+x+1)^2  = x^2 + 4x + 1 / (x^2+x+1)^2 . 
f'(x) | x= 0  = 1

由題意,拋物線 y = ax^2+bx+c 與直線 7x - y - 8 =0相切,故可視7x - y - 8 =0為其切線,切點為(2,6) 
7x - y - 8 = 0 
7x - 8 = y , m = 7 
m = y' | x =2 , y' = 2ax+ b 將 x = 2 代入,  7 = 4a + b ---- (1).  
(2,6) 為  y = ax^2+bx+c 的切點 ,則 6 = 4a + 2b + c ---- (2)
(2) -(1) : b+c = -1 
又與 x-y+1= 0 相切. -y = -x -1 , y= x+ 1故另條切線的斜率為 m' = 1, 其切點
2ax + b = 1  , 2ax  = 1 -b , x = (1- b)/2a 將代入     (1- b)/2a - y + 1 = 0 , y  =  (1- b)/2a +1  = 2a - b +1 / 2a  

這種題目要注意, 因為曲線為三次多項式. 這裡 有一個詳細討論. 
假設切點為(t,t^3-3t^2) , why ? 因為此切點必為曲線Γ的一點 . 
y = x^3 -3x^2 , y ' = 3x^2 - 6x 
m = y' | x = a = 3a^2 -6a 因為此切線也過P(2,-5) 所以, 此切線方程式可表示如下: y - (-5) = m (x-2) 
      y + 5 = (3a^2 - 6a)(x-2) 再將P帶入得: