The Differentiation of the Inverse Functions Differential and Linear Approximation # The Differentiation of the Inverse Functions# The idea of Inverse FunctionSuppose y = f ( x ) = 3 x + 5 y = f(x) = 3x+5 y = f ( x ) = 3 x + 5 ⇒ x = 1 3 ( y − 5 ) \Rightarrow x = \frac{1}{3}(y-5) ⇒ x = 3 1 ( y − 5 ) ⇒ y = 1 3 ( x − 5 ) \Rightarrow y = \frac{1}{3}(x-5) ⇒ y = 3 1 ( x − 5 ) f − 1 ( x ) = 1 3 ( x − 5 ) f^{-1}(x) = \frac{1}{3}(x-5) f − 1 ( x ) = 3 1 ( x − 5 ) is the inverse function of f ( x ) = 3 x + 5 f(x) = 3x+5 f ( x ) = 3 x + 5
Consider y = f ( x ) = e x y = f(x) = e^x y = f ( x ) = e x ⇒ l o g e y = l o g e e x = x \Rightarrow log_ey = log_e e^x = x ⇒ l o g e y = l o g e e x = x ⇒ x = l o g e y \Rightarrow x = log_ey ⇒ x = l o g e y ⇒ y = l o g e x \Rightarrow y = log_ex ⇒ y = l o g e x f − 1 ( x ) = l o g e x f^{-1}(x) = log_ex f − 1 ( x ) = l o g e x is the inverse function of f ( x ) = e x f(x) = e^x f ( x ) = e x
f ( x ) = s i n − 1 x ⇒ f : [ − 1 : 1 ] → [ − π 2 , π 2 ] f(x) = sin^{-1}x \; \Rightarrow \; f:[-1:1]\to [ -\frac{\pi}{2},\frac{\pi}{2}] f ( x ) = s i n − 1 x ⇒ f : [ − 1 : 1 ] → [ − 2 π , 2 π ] f ( x ) = c o s − 1 x ⇒ f : [ − 1 : 1 ] → [ 0 , π ] f(x) = cos^{-1}x \; \Rightarrow \; f:[-1:1]\to [ 0 , \pi ] f ( x ) = c o s − 1 x ⇒ f : [ − 1 : 1 ] → [ 0 , π ] f ( x ) = t a n − 1 x ⇒ f : R → [ − π 2 , π 2 ] f(x) = tan^{-1}x \; \Rightarrow \; f: \mathbb{R} \to [ -\frac{\pi}{2},\frac{\pi}{2}] f ( x ) = t a n − 1 x ⇒ f : R → [ − 2 π , 2 π ] f ( x ) = c o t − 1 x ⇒ f : R → [ 0 , π ] f(x) = cot^{-1}x \; \Rightarrow \; f: \mathbb{R} \to [ 0 , \pi ] f ( x ) = c o t − 1 x ⇒ f : R → [ 0 , π ] f ( x ) = s e c − 1 x ⇒ f : ( − ∞ , − 1 ] ∪ [ 1 , ∞ ) → [ 0 , π 2 ) ∪ ( π 2 , π ] f(x) = sec^{-1}x \; \Rightarrow \; f: (-\infty,-1] \cup [1,\infty) \to [ 0 , \frac{\pi}{2}) \cup (\frac{\pi}{2},\pi] f ( x ) = s e c − 1 x ⇒ f : ( − ∞ , − 1 ] ∪ [ 1 , ∞ ) → [ 0 , 2 π ) ∪ ( 2 π , π ] f ( x ) = c s c − 1 x ⇒ f : ( − ∞ , − 1 ] ∪ [ 1 , ∞ ) → [ − π 2 , 0 ) ∪ ( 0 , π 2 ] f(x) = csc^{-1}x \; \Rightarrow \; f: (-\infty,-1] \cup [1,\infty) \to [ -\frac{\pi}{2} , 0 ) \cup ( 0, \frac{\pi}{2}] f ( x ) = c s c − 1 x ⇒ f : ( − ∞ , − 1 ] ∪ [ 1 , ∞ ) → [ − 2 π , 0 ) ∪ ( 0 , 2 π ] Remark domain of f f f = range of f − 1 f^{-1} f − 1 . range of f f f = domain of f − 1 f^{-1} f − 1 . y = f ( x ) ⇄ x = f − 1 ( y ) y=f(x) \rightleftarrows x = f^{-1}(y) y = f ( x ) ⇄ x = f − 1 ( y ) f ( f − 1 ( y ) ) = f ( x ) = y f(f^{-1}(y)) = f(x) = y f ( f − 1 ( y ) ) = f ( x ) = y f − 1 ( f ( x ) ) = f − 1 ( y ) = x f^{-1}(f(x)) = f^{-1}(y) = x f − 1 ( f ( x ) ) = f − 1 ( y ) = x f f f 和f − 1 f^{-1} f − 1 的圖形對稱於y = x y=x y = x # The Differentiation of the Inverse FunctionsSuppose f f f is differentiable and its inverse function exists. Let y = f − 1 ( x ) y = f^{-1}(x) y = f − 1 ( x ) and then x = f ( y ) x = f(y) x = f ( y ) Since f ( f − 1 ( x ) ) = x f(f^{-1}(x)) = x f ( f − 1 ( x ) ) = x
∵ d d x { f ( f − 1 ( x ) ) } = f ′ ( f − 1 ( x ) ) ⋅ d d x f − 1 ( x ) = d d x { x } = 1 \because \frac{d}{dx} \{ f(f^{-1}(x)) \} = f'(f^{-1}(x)) \cdot \frac{d}{dx}{f^{-1}(x)} = \frac{d}{dx}\{ x \} = 1 ∵ d x d { f ( f − 1 ( x ) ) } = f ′ ( f − 1 ( x ) ) ⋅ d x d f − 1 ( x ) = d x d { x } = 1 ⇒ f ′ ( f − 1 ( x ) ) ⋅ d d x f − 1 ( x ) = 1 \Rightarrow f'(f^{-1}(x)) \cdot \frac{d}{dx}f^{-1}(x) = 1 ⇒ f ′ ( f − 1 ( x ) ) ⋅ d x d f − 1 ( x ) = 1 ⇒ d d x f ′ ( x ) = 1 f ′ ( f − 1 ( x ) ) = 1 f ′ ( y ) = 1 d d y f ( y ) \Rightarrow \frac{d}{dx}f'(x) = \frac{1}{f'(f^{-1}(x))} = \frac{1}{f'(y)} = \frac{1}{\frac{d}{dy}f(y)} ⇒ d x d f ′ ( x ) = f ′ ( f − 1 ( x ) ) 1 = f ′ ( y ) 1 = d y d f ( y ) 1 if f ′ ( y ) ≠ 0 f'(y) \neq 0 f ′ ( y ) = 0
Let f ( x ) = 1 4 x 3 + x − 1 f(x) = \frac{1}{4}x^3+x-1 f ( x ) = 4 1 x 3 + x − 1
What is the value of f − 1 ( x ) f^{-1}(x) f − 1 ( x ) when x = 3 x = 3 x = 3 ? What is the value of ( f − 1 ) ′ ( x ) (f^{-1})'(x) ( f − 1 ) ′ ( x ) when x = 3 x = 3 x = 3 ? Let f ( x ) = 3 ⇒ x 3 + 4 x − 16 = 0 f(x) = 3 \Rightarrow x^3 + 4x - 16 = 0 f ( x ) = 3 ⇒ x 3 + 4 x − 1 6 = 0 ⇒ ( x − 2 ) ( x 2 + 2 x + 8 ) = 0 ∴ x = 2 ⇒ f − 1 ( 3 ) = 2 \Rightarrow (x-2)(x^2+2x+8) = 0 \therefore x = 2 \Rightarrow f^{-1}(3) = 2 ⇒ ( x − 2 ) ( x 2 + 2 x + 8 ) = 0 ∴ x = 2 ⇒ f − 1 ( 3 ) = 2
d d x ∣ x = 3 f − 1 ( x ) = 1 d d y ∣ y = f − 1 ( 3 ) f ( y ) = 1 ( 3 4 y 2 + 1 ) ∣ y = 2 = 1 4 \frac{d}{dx}|_{x=3}f^{-1}(x) = \frac{1}{\frac{d}{dy}|_{y=f^{-1}(3)}f(y)} = \frac{1}{(\frac{3}{4}y^2+1)|_{y=2}} = \frac{1}{4} d x d ∣ x = 3 f − 1 ( x ) = d y d ∣ y = f − 1 ( 3 ) f ( y ) 1 = ( 4 3 y 2 + 1 ) ∣ y = 2 1 = 4 1 .
y = f − 1 ( x ) = s i n − 1 x y = f^{-1}(x) = sin^{-1}x y = f − 1 ( x ) = s i n − 1 x d d x f − 1 ( x ) = d d x s i n − 1 x = 1 d d y f ( y ) = 1 c o s y = 1 1 − x 2 \frac{d}{dx} f^{-1}(x) = \frac{d}{dx} sin^{-1}x = \frac{1}{\frac{d}{dy}f(y)}=\frac{1}{cosy} = \frac{1}{\sqrt{1-x^2}} d x d f − 1 ( x ) = d x d s i n − 1 x = d y d f ( y ) 1 = c o s y 1 = 1 − x 2 1
d d x e x = e x \frac{d}{dx}e^x = e^x d x d e x = e x d d x a x = ( l n a ) ⋅ a x \frac{d}{dx}a^x = (lna) \cdot a^x d x d a x = ( l n a ) ⋅ a x d d x ( l o g a x ) = 1 d d y a y = 1 ( l n a ) a y = 1 x l n a \frac{d}{dx}(log_ax) = \frac{1}{\frac{d}{dy}a^y} = \frac{1}{(lna)a^y} = \frac{1}{xlna} d x d ( l o g a x ) = d y d a y 1 = ( l n a ) a y 1 = x l n a 1 d d x l n x = 1 x , x > 0 \frac{d}{dx}lnx = \frac{1}{x},x>0 d x d l n x = x 1 , x > 0 d d x x n = n x n − 1 \frac{d}{dx}x^n = nx^{n-1} d x d x n = n x n − 1 if n ∈ R n\in \mathbb{R} n ∈ R Let e = lim n → ∞ ( 1 + 1 n ) n e = \lim_{n \to \infty}(1+\frac{1}{n})^n e = lim n → ∞ ( 1 + n 1 ) n Then e x = lim n → ∞ ( 1 + x n ) n e^x = \lim_{n\to \infty}(1+\frac{x}{n})^n e x = lim n → ∞ ( 1 + n x ) n # Differential and Linear Approximation# DefineSuppose f f f is differentiable at a a a .Then when x x x is closed to a a a .f f f looks like its tangent line y = f ( a ) + f ′ ( a ) ( x − a ) y = f(a) + f'(a)(x-a) y = f ( a ) + f ′ ( a ) ( x − a ) and we have f ( x ) ≈ f ( a ) + f ′ ( a ) ( x − a ) f(x) \approx f(a) + f'(a)(x-a) f ( x ) ≈ f ( a ) + f ′ ( a ) ( x − a ) .f ( a ) + f ′ ( a ) ( x − a ) f(a) + f'(a)(x-a) f ( a ) + f ′ ( a ) ( x − a ) is called linear approximation of f f f at a a a , 記為L ( x ) = f ( a ) + f ′ ( a ) ( x − a ) L(x) = f(a) +f'(a)(x-a) L ( x ) = f ( a ) + f ′ ( a ) ( x − a ) .
# DefineIf f f f is defined by the equation y = f ( x ) y = f(x) y = f ( x ) ,then the differential of y y y ,denoted by d y dy d y , is given by d y = f ′ ( x ) Δ x dy = f'(x)\Delta x d y = f ′ ( x ) Δ x where Δ x \Delta x Δ x is an arbitrary invrement of x x x .
If f f f is defined by the equation y = f ( x ) y = f(x) y = f ( x ) ,then the differential of x x x ,denoted by d x dx d x , is given by d x = Δ x dx = \Delta x d x = Δ x where Δ x \Delta x Δ x is an arbitrary invrement of x x x .
由以上定義,可得d y = f ′ ( x ) d x dy = f'(x)dx d y = f ′ ( x ) d x ∵ Δ y = f ( x + Δ x ) − f ( x ) \because \Delta y = f(x+\Delta x) -f(x) ∵ Δ y = f ( x + Δ x ) − f ( x ) (Δ y \Delta y Δ y is the increment of y y y )∴ Δ y ≈ d y \therefore \Delta y \approx dy ∴ Δ y ≈ d y when Δ x \Delta x Δ x is very small.
# Reference蘇承芳老師 - 微積分甲(一)109 學年度 - Calculus (I) Academic Year 109