Matematické Fórum

stuart clark · 20. 08. 2011 05:56

If $f(x) = \left(2x-\pi\right)^3+2x-\cos x$ . then $\frac{d}{dx}\left(f^{-1}(x)\right)\Bigint|_{x=\pi}$

jarrro · 20. 08. 2011 11:06

Pavel Brožek · 20. 08. 2011 12:38

↑ jarrro:

I think your result is $\frac{d}{dx}\left(f^{-1}(x)\right)\Bigint|_{x=f(\pi)}$ .

jarrro · 20. 08. 2011 19:06 — Editoval jarrro (20. 08. 2011 19:12)

↑ Pavel Brožek:I use formula $f^{-1}^{\prime}{\left(x\right)}=\frac{1}{f^{\prime}{\left(f{\left(x\right)}\right)}}$
and substitute x=pi

Pavel Brožek · 20. 08. 2011 21:02

↑ jarrro:

But the correct formula is $f^{-1}^{\prime}{\left(x\right)}=\frac{1}{f^{\prime}{\left(f^{-1}{\left(x\right)}\right)}}$ .

jarrro · 20. 08. 2011 21:11 — Editoval jarrro (20. 08. 2011 21:35)

↑ Pavel Brožek:I am an idiot sorry
but I hope that we can write
$f^{-1}^{\prime}{\left(f{\left(x\right)}\right)}=\frac{1}{f^{\prime}{\left(x\right)}}$
and to $\frac{1}{6\left(2x-\pi\right)^2+2+\sin{x}}$ substitute the solution of equation
$\left(2x-\pi\right)^3+2x-\cos{x}=\pi$

Matematické Fórum

#1 20. 08. 2011 05:56

diff. of inverse

#2 20. 08. 2011 11:06

Re: diff. of inverse

#3 20. 08. 2011 12:38

Re: diff. of inverse

#4 20. 08. 2011 19:06 — Editoval jarrro (20. 08. 2011 19:12)

Re: diff. of inverse

#5 20. 08. 2011 21:02

Re: diff. of inverse

#6 20. 08. 2011 21:11 — Editoval jarrro (20. 08. 2011 21:35)

Re: diff. of inverse

Zápatí