Re: [考古] 91交大

[goshfju (cola)]

※ 引述《kusorz (^~^)》之銘言：
: suppose the f is one to one and f(1)=2,f(2)=3,f'(1)=1,f'(2)=3,f'(3)=5
: find (f^-1)'(2)=?

          -1
f(1)=2 , f  (2)=1

  -1              -1
(f  )'(2) = 1/f'(f  (2)) = 1/f'(1) = 1/1 = 1

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推 MrTang:我算1/2...真糗~~~
→ MrTang:能問一下f'(2)的反函數~下一步是怎麼來的.
推 MrTang:想不太出來
→ goshfju:你是指反函數微分的公式嗎?
→ goshfju:其實上面3261篇就有板友貼了 你可以去參考
→ goshfju: y=f^-1 => f(y)=x => f(f^-1(x))=x
→ goshfju:上式對x微分 f'(f^-1(x))*(f^-1)'(x)=1
→ goshfju:(f^-1)'(x) = 1 / f'(f^-1(x))
→ MrTang:謝謝~