How Is the Second Derivative of an Inverse Function Calculated?

[Jhenrique]

I can derivate x(y) wrt y using the derivative of y(x) wrt x, follows the formula: $$\frac{dx}{dy}=\frac{1}{\frac{dy}{dx}}$$ until same the 2nd derivative (taking the 2nd diff form of x and deriving wrt to x):$$d^2x=\frac{d^2 x}{dy^2} dy^2 + \frac{dx}{dy} d^2y$$ $$\frac{d^2x}{dx^2}=\frac{d^2 x}{dy^2} \frac{dy^2}{dx^2} + \frac{dx}{dy} \frac{d^2y}{dx^2}$$ $$0=\frac{d^2 x}{dy^2} \frac{dy^2}{dx^2} + \frac{dx}{dy} \frac{d^2y}{dx^2}$$ solving for d²x/dy²: $$\frac{d^2x}{dy^2}=-\frac{d^2y}{dx^2}\frac{1}{\left( \frac{dy}{dx} \right)^3}$$ I think that this is a razoable deduction for the formula of d²x/dy² in terms of derivatives of y wrt x.


Now, where it came from this formula?

links:
http://en.wikipedia.org/wiki/Integration_of_inverse_functions
http://en.wikipedia.org/wiki/Inverse_functions_and_differentiation

 

[Boson]

Put y=f(x) and solve by substitution method.

 

[Jhenrique]

The answer is here: https://en.wikipedia.org/wiki/Integration_by_parts#Visualization

 

[Mark44]

I can derivate x(y) wrt y using the derivative of y(x) wrt x, follows the formula: $$\frac{dx}{dy}=\frac{1}{\frac{dy}{dx}}$$ until same the 2nd derivative (taking the 2nd diff form of x and deriving wrt to x):$$d^2x=\frac{d^2 x}{dy^2} dy^2 + \frac{dx}{dy} d^2y$$ $$\frac{d^2x}{dx^2}=\frac{d^2 x}{dy^2} \frac{dy^2}{dx^2} + \frac{dx}{dy} \frac{d^2y}{dx^2}$$ $$0=\frac{d^2 x}{dy^2} \frac{dy^2}{dx^2} + \frac{dx}{dy} \frac{d^2y}{dx^2}$$ solving for d²x/dy²: $$\frac{d^2x}{dy^2}=-\frac{d^2y}{dx^2}\frac{1}{\left( \frac{dy}{dx} \right)^3}$$ I think that this is a razoable deduction for the formula of d²x/dy² in terms of derivatives of y wrt x.

I don't know if the above is reasonable (no such word as "razoable" in English). It would help if you added some words to say what you're doing.

BTW, you don't "derivate" something to get its derivative - you differentiate it. As far as I know, "derivate" is not a word, certainly not as used in mathematics.

Now, where it came from this formula?

There seems to be an error in this formula.
It should say the following:
$$ \int f^{-1}(y)dy = yf^{-1}(y) - F ° f^{-1}(y) + C$$
The error in the wiki page you linked to is in the first term on the right side. They have x f^-1(y), but it should by yf^-1(y). The three examples are consistent with what I have, but not what they have for their general formula.

Calculating this integral is not very difficult using integration by parts.

Let y = f(x) (⇔ x = f^-1(y))
Then dy = f'(x) dx

Using this substitution, the integral $\int f^{-1}(y)dy$ becomes $\int x \cdot f'(x)dx$.

Using integration by parts, with u = x and dv = f'(x)dx,
we have du = dx and v = ∫f'(x)dx = f(x). (Don't need the constant yet.)

So $\int x \cdot f'(x)dx = x f(x) - \int f(x) dx$
Since x = f^-1(y), and f(x) = y, we have
f^-1(y) * y + F(x) + C

Here, F is an antiderivative of f.

links:
http://en.wikipedia.org/wiki/Integration_of_inverse_functions
http://en.wikipedia.org/wiki/Inverse_functions_and_differentiation