prove-that-df-1-a-dx-df-f-1-a-dx-1-

Question Number 166331 by mr W last updated on 18/Feb/22

p r o v e t h a t

\frac{{d f}^{- 1} (a)}{d x} \times \frac{d f (f^{- 1} (a))}{d x} = 1

Commented by mr W last updated on 19/Feb/22

i t i s m e a n t t h a t

\frac{{d f}^{- 1} (a)}{d x} = \frac{{d f}^{- 1} (x)}{d x} ∣_{x = a}

\frac{d f (f^{- 1} (a))}{d x} = \frac{d f (x)}{d x} ∣_{x = f^{- 1} (a)}

Answered by mathsmine last updated on 18/Feb/22

{f o f}^{-} (x) = x

\Rightarrow \frac{{d f}^{-} (x)}{d x} . \frac{d f}{d x} (f^{-} (x)) = 1

x = a

Commented by mr W last updated on 18/Feb/22

t h a n k s s i r!

Answered by mr W last updated on 19/Feb/22

Commented by mr W last updated on 18/Feb/22

g e o m e t r i c w a y :

f^{- 1} (x) a n d f (x) a r e s y m m e t r i c a b o u t

t h e l i n e y = x .

t a n g e n t l i n e a t Q (a, f^{- 1} (a)) i s

s y m m e t r i c t o t h e t a n g e n t l i n e a t

P (f^{- 1} (a), a) .

\frac{{d f}^{- 1} (a)}{d x} = \tan (\frac{π}{2} - θ) = \frac{1}{\tan θ}

\frac{d f (f^{- 1} (a))}{d x} = \tan θ

\frac{{d f}^{- 1} (a)}{d x} \times \frac{d f (f^{- 1} (a))}{d x} = \frac{1}{\tan θ} \times \tan θ = 1

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