I need help finding the derivative of an inverse function

logan.sowder38

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\(\displaystyle 1.\, \mbox{ If }\, f(x)\, =\, \dfrac{x^3}{x^2\, +\, 1},\, \mbox{ find }\, \left(f^{-1}\right)' (2)\)

Here's my work so far.

\(\displaystyle f(x)\, =\, \dfrac{x^3}{x^2\, +\, 1}\). . . . .\(\displaystyle \mbox{Find }\, \left(f^{-1}\right)' (2)\)

\(\displaystyle \mbox{Let }\, g(x)\, =\, f^{-1}(x)\, \longrightarrow \, \mbox{ find }\, g'(2)\)

\(\displaystyle g'(x)\, =\, \dfrac{1}{f'(g(x))}\)


\(\displaystyle f(x)\, =\, 2\)

\(\displaystyle 2\, =\, \dfrac{x^3}{x^2\, +\, 1}\)

\(\displaystyle 2x^2\, +\, 2\, =\, x^3\)



\(\displaystyle x^3\, -\, 2x^2\, -\, 2\, =\, 0\, ???\)
\(\displaystyle f'(x)\, =\, \dfrac{(x^2\, +\, 1)(3x^2)\, -\, (x^3)(2x)}{(x^2\, +\, 1)^2}\)

\(\displaystyle f'(x)\, =\, \dfrac{3x^4\, +\, 3x^2\, -\, 6x^3}{(x^2\, +\, 1)^2}\)

\(\displaystyle f'(x)\, =\, \dfrac{3x^4\, -\, 6x^3\, +\, 3x^2}{x^4\, +\, 2x^2\, +\, 1}\)

I understand to use the theorem (f-1)' (a) = 1/(f'(f-1(a)), but I can't figure out how to solve for x so I can plug it into f'(x)
 
Last edited by a moderator:
\(\displaystyle 1.\, \mbox{ If }\, f(x)\, =\, \dfrac{x^3}{x^2\, +\, 1},\, \mbox{ find }\, \left(f^{-1}\right)' (2)\)

Here's my work so far.

\(\displaystyle f(x)\, =\, \dfrac{x^3}{x^2\, +\, 1}\). . . . .\(\displaystyle \mbox{Find }\, \left(f^{-1}\right)' (2)\)

\(\displaystyle \mbox{Let }\, g(x)\, =\, f^{-1}(x)\, \longrightarrow \, \mbox{ find }\, g'(2)\)

\(\displaystyle g'(x)\, =\, \dfrac{1}{f'(g(x))}\)


\(\displaystyle f(x)\, =\, 2\)

\(\displaystyle 2\, =\, \dfrac{x^3}{x^2\, +\, 1}\)

\(\displaystyle 2x^2\, +\, 2\, =\, x^3\)



\(\displaystyle x^3\, -\, 2x^2\, -\, 2\, =\, 0\, ???\)
\(\displaystyle f'(x)\, =\, \dfrac{(x^2\, +\, 1)(3x^2)\, -\, (x^3)(2x)}{(x^2\, +\, 1)^2}\)

\(\displaystyle f'(x)\, =\, \dfrac{3x^4\, +\, 3x^2\, -\, 6x^3}{(x^2\, +\, 1)^2}\)<== -2x4 not -6x3

\(\displaystyle f'(x)\, =\, \dfrac{3x^4\, -\, 6x^3\, +\, 3x^2}{x^4\, +\, 2x^2\, +\, 1}\)

I understand to use the theorem (f-1)' (a) = 1/(f'(f-1(a)), but I can't figure out how to solve for x so I can plug it into f'(x)

I would think there would be an easier solution but I don't see it. You just have to solve
\(\displaystyle x^3\, -\, 2x^2\, -\, 2\, =\, 0\)
Note that you can write f' as [see the note in red above]
\(\displaystyle f'(x)\, =\, \dfrac{x^2\, +\, 3}{x^4}\, f^2(x)\)
or several other equivalent ways but I don't see that as any real help.



An iterative solution which should work pretty fast is
\(\displaystyle x_{i+1}\, =\, \frac{2}{x_i^2}\, +\, 2\)
where xi is the ith iterate. A good starting value would be x0=2.3
 
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