1f(a)=Ï€
x=Ï€2
f'(Ï€2)=3
now use
f^{-1}(f(x))=x \\ \frac{d }{d x}{f^{-1}(f(x))}=1\\ \texttt{apply chain rule} \\ \frac{d }{d x}{f^{-1}(f(a))}=\frac{1}{f'(a)}
\frac{d }{d x}{f^{-1}(\pi)}=\frac{1}{3}
I could not solve this for quite sometime...
lets see who can ...
let f(x)=(2x-Ï€)3 + 2x - cos x
find ddx[f-1(x)] at x=Ï€ [1][1][1]
-
UP 0 DOWN 0 0 3
3 Answers
1hey subho.....
just check out this link 4 da solution......
http://www.targetiit.com/iit-jee-forum/posts/differentiation-2-17388.html
i had also asked the same doubt earlier.....
1
49no i have solved this now...
just gave this as i found the question fascinating!
neways thanks! :)