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Suppose f is differentiable with derivative f '(x) = (1+x^3) ^(1/2).

If g = f^-1, show that g ''(x) = 3/2 g(x)^2.

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Suppose f is differentiable with derivative f '(x) = (1+x^3) ^(1/2).

If g = f^-1, show that g ''(x) = 3/2 g(x)^2.

- #2

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Write f(x) = int[0,x, (1+t^3) ^(1/2),dt]

Set the integral equal to u so that you have:

f(x) = u

Take the inverse of both sides:

x = g(u)

Differentiate wrt x, using the chain rule:

(dg/du)(du/dx) = 1

Using FTC, du/dx = (1+x^3) ^(1/2)

Then dg/du = (1+x^3) ^(-1/2)

but remember x =g(u)

So g' = (1+g^3) ^(-1/2)

differentiate again (remember to use the chain rule):

g'' = (-1/2)(3g)(g')(1+g^3) ^(-3/2)

g'' = (-1/2)(3g)(1+g^3) ^(-1/2))(1+g^3) ^(-3/2)

Which simplifies to your expression. I am sorry to have deprived you of solving this wonderful problem, but it is pretty difficult to get started on a problem like this if you do not know where to begin.

I invented this method to to derive an expression for the integral int[0,x, 1/t, dt] .

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wow, thank you so much, you are amazing

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mathwonk

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If g = f^-1, show that g ''(x) = 3/2 g(x)^2. \\

this seems incorrect, as shown by the previous post, which also seems however slightly incorrect.

i.e. dx/dy = 1/(dy/dx). so dx/dy = (1+x^3)^(-1/2). so d^2x/dy^2, by chain rule,

= (-1/2)(1+x^3)^(-3/2) (3x^2) dx/dy = (-1/2)(1+x^3)^(-3/2) (3x^2)(1+x^3)^(-1/2)

= (-1/2) (1+x^3)^(-2) (3x^2)

= (-3/2)[1+g(y)^3)^(-2)] (g(y)^2). (the previous post omitted the power 2.)

in particular all the integration in the previous post is entirely superfluous.

now i may easily be missing something here, but my problem is that now I do not see how this simplifies to (3/2) [g(y)]^2, since it does not appear to me that

[1+g(y)^3)^(-2)] = -1, i.e. that

1 + x^3 = -1, i.e.that x = cuberoot(-2), for all x.

what am i missing?

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matt grime

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mathwonk

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I probably meant [1+g(y)^3)]^(-2) = -1, so the left side is a square while the right side is negative.

the point is the problem was probably miscopied from the source.

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matt grime

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mathwonk

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