• The Differentiation of the Inverse Functions
  • Differential and Linear Approximation

# The Differentiation of the Inverse Functions

# The idea of Inverse Function

Suppose y=f(x)=3x+5y = f(x) = 3x+5
⇒x=13(y−5)\Rightarrow x = \frac{1}{3}(y-5)
⇒y=13(x−5)\Rightarrow y = \frac{1}{3}(x-5)
f−1(x)=13(x−5)f^{-1}(x) = \frac{1}{3}(x-5) is the inverse function of f(x)=3x+5f(x) = 3x+5

Consider y=f(x)=exy = f(x) = e^x
⇒logey=logeex=x\Rightarrow log_ey = log_e e^x = x
⇒x=logey\Rightarrow x = log_ey
⇒y=logex\Rightarrow y = log_ex
f−1(x)=logexf^{-1}(x) = log_ex is the inverse function of f(x)=exf(x) = e^x

  1. f(x)=sin−1x⇒f:[−1:1]→[−π2,π2]f(x) = sin^{-1}x \; \Rightarrow \; f:[-1:1]\to [ -\frac{\pi}{2},\frac{\pi}{2}]
  2. f(x)=cos−1x⇒f:[−1:1]→[0,π]f(x) = cos^{-1}x \; \Rightarrow \; f:[-1:1]\to [ 0 , \pi ]
  3. f(x)=tan−1x⇒f:R→[−π2,π2]f(x) = tan^{-1}x \; \Rightarrow \; f: \mathbb{R} \to [ -\frac{\pi}{2},\frac{\pi}{2}]
  4. f(x)=cot−1x⇒f:R→[0,π]f(x) = cot^{-1}x \; \Rightarrow \; f: \mathbb{R} \to [ 0 , \pi ]
  5. f(x)=sec−1x⇒f:(−∞,−1]∪[1,∞)→[0,π2)∪(π2,π]f(x) = sec^{-1}x \; \Rightarrow \; f: (-\infty,-1] \cup [1,\infty) \to [ 0 , \frac{\pi}{2}) \cup (\frac{\pi}{2},\pi]
  6. f(x)=csc−1x⇒f:(−∞,−1]∪[1,∞)→[−π2,0)∪(0,π2]f(x) = csc^{-1}x \; \Rightarrow \; f: (-\infty,-1] \cup [1,\infty) \to [ -\frac{\pi}{2} , 0 ) \cup ( 0, \frac{\pi}{2}]
Remark
  1. domain of ff = range of f−1f^{-1}.
    range of ff = domain of f−1f^{-1}.
  2. y=f(x)⇄x=f−1(y)y=f(x) \rightleftarrows x = f^{-1}(y)
  3. f(f−1(y))=f(x)=yf(f^{-1}(y)) = f(x) = y
    f−1(f(x))=f−1(y)=xf^{-1}(f(x)) = f^{-1}(y) = x
  4. ff 和f−1f^{-1} 的圖形對稱於y=xy=x

# The Differentiation of the Inverse Functions

Suppose ff is differentiable and its inverse function exists.
Let y=f−1(x)y = f^{-1}(x) and then x=f(y)x = f(y)
Since f(f−1(x))=xf(f^{-1}(x)) = x

∵ddx{f(f−1(x))}=f′(f−1(x))⋅ddxf−1(x)=ddx{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
⇒f′(f−1(x))⋅ddxf−1(x)=1\Rightarrow f'(f^{-1}(x)) \cdot \frac{d}{dx}f^{-1}(x) = 1
⇒ddxf′(x)=1f′(f−1(x))=1f′(y)=1ddyf(y)\Rightarrow \frac{d}{dx}f'(x) = \frac{1}{f'(f^{-1}(x))} = \frac{1}{f'(y)} = \frac{1}{\frac{d}{dy}f(y)} if f′(y)≠0f'(y) \neq 0

Let f(x)=14x3+x−1f(x) = \frac{1}{4}x^3+x-1

  1. What is the value of f−1(x)f^{-1}(x) when x=3x = 3 ?
  2. What is the value of (f−1)′(x)(f^{-1})'(x) when x=3x = 3 ?

Let f(x)=3⇒x3+4x−16=0f(x) = 3 \Rightarrow x^3 + 4x - 16 = 0
⇒(x−2)(x2+2x+8)=0∴x=2⇒f−1(3)=2\Rightarrow (x-2)(x^2+2x+8) = 0 \therefore x = 2 \Rightarrow f^{-1}(3) = 2

ddx∣x=3f−1(x)=1ddy∣y=f−1(3)f(y)=1(34y2+1)∣y=2=14\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}.

  1. 反三角函數
  2. 指對數函數

y=f−1(x)=sin−1xy = f^{-1}(x) = sin^{-1}x
ddxf−1(x)=ddxsin−1x=1ddyf(y)=1cosy=11−x2\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}}

  1. ddxex=ex\frac{d}{dx}e^x = e^x
  2. ddxax=(lna)⋅ax\frac{d}{dx}a^x = (lna) \cdot a^x
  3. ddx(logax)=1ddyay=1(lna)ay=1xlna\frac{d}{dx}(log_ax) = \frac{1}{\frac{d}{dy}a^y} = \frac{1}{(lna)a^y} = \frac{1}{xlna}
    ddxlnx=1x,x>0\frac{d}{dx}lnx = \frac{1}{x},x>0
  4. ddxxn=nxn−1\frac{d}{dx}x^n = nx^{n-1}if n∈Rn\in \mathbb{R}
  5. Let e=lim⁡n→∞(1+1n)ne = \lim_{n \to \infty}(1+\frac{1}{n})^n Then ex=lim⁡n→∞(1+xn)ne^x = \lim_{n\to \infty}(1+\frac{x}{n})^n

# Differential and Linear Approximation

# Define

Suppose ff is differentiable at aa.Then when xx is closed to aa.
ff looks like its tangent line 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(a)+f′(a)(x−a)f(a) + f'(a)(x-a) is called linear approximation of ff at aa, 記為L(x)=f(a)+f′(a)(x−a)L(x) = f(a) +f'(a)(x-a).

# Define

If ff is defined by the equation y=f(x)y = f(x),then the differential of yy,denoted by dydy, is given by dy=f′(x)Δxdy = f'(x)\Delta x where Δx\Delta x is an arbitrary invrement of xx.

If ff is defined by the equation y=f(x)y = f(x),then the differential of xx,denoted by dxdx, is given by dx=Δxdx = \Delta x where Δx\Delta x is an arbitrary invrement of xx.

由以上定義,可得dy=f′(x)dxdy = f'(x)dx
∵Δy=f(x+Δx)−f(x)\because \Delta y = f(x+\Delta x) -f(x)(Δy\Delta y is the increment of yy)
∴Δy≈dy\therefore \Delta y \approx dy when Δx\Delta x is very small.


# Reference

  • 蘇承芳老師 - 微積分甲(一)109 學年度 - Calculus (I) Academic Year 109
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