Question

Simplest possible derivation for\[\frac{d}{dx}f^{-1}(x)=\frac{1}{f'[f^{-1}(x)]}\]please.

Answer 1

I found a few geometric-type proofs, but I just want to do it analytically.

Answer 2

Let f be a bijective application and assume its reciprocal application can be differentiated in a point d=f(c) where c is a point where f can also be differentiated
Let's name g our reciprocal function, then
\[g=f^{-1}, f(c)=d\]
\[\frac{g(y)-g(d)}{y-d}=\frac{x-c}{f(x)-f(c)}\]
and
\[\lim_{x \rightarrow c}\frac{x-c}{f(x)-f(c)}=\frac{1}{f'(c)}\]
but
\[f(c)=d \iff f^{-1}(d)=c\]
because f is bijective
then
\[\lim_{x \rightarrow c} \frac{x-c}{f(x)-f(c)}=\lim_{y \rightarrow d} \frac{g(y)-g(d)}{y-d}=\frac{1}{f'(c)}=\frac{1}{f'(f^{-1}(d)}\]

Answer 3

Great, thanks.
Any others proofs would be appreciated as well guys!

Answer 4

This is a pseudoproof, it's not really rigorous but here it goes..\[f(f^{-1}(x))=x \Rightarrow f'(f^{-1}(x))*f'^{-1}(x) =1 \Rightarrow f'^{-1}(x)=\frac{1}{f'(f^{-1}(x)}\]

Answer 5

Even better, that pseudo-proof will be easier to call to mind. Wish I could give you two medals ;-)

Answer 6

nooo problem :p