Is there any pattern for the derivative of $y'$ with respect to $y$? [closed]

How would you calculate $\frac{d\frac{d y}{d x}}{d y}$? In addition, how would you calculate $\int y' \, dy$?

Solution 1:

As for the integral, we can calculate the derivative of $y'$ with respect to $y$

Let's define $f(x)=y$ and $g(y)=y'(x(y))$ being $x(y)=f^{-1}(y)$. Then, by the chain rule,

$\dfrac{dg}{dy}=\dfrac{d}{dx}\left(\dfrac{df}{dx}\right)\dfrac{dx}{dy}=$

$=\dfrac{d^2f}{dx^2}\dfrac{df^{-1}}{dy}=y''\dfrac{df^{-1}}{dy}$

and finally, by setting $h(y)=\dfrac{d^2f}{dx^2}(x(y))$, $\dfrac{dg}{dy}=h(y)\dfrac{df^{-1}}{dy}$

But you can calculate that derivative directly on $g(y)$

E.g.:

$y=f(x)=\ln(x)$, $x=f^{-1}(y)=e^y$

$g(y)=y'(x)=1/x=1/e^y$, $\dfrac{dg}{dy}=-1/e^y$

Or

With $y''(x)=-1/x^2=-1/e^{2y}$ and $\dfrac{df^{-1}}{dy}=e^y$

$\dfrac{dg}{dy}=-\dfrac{1}{e^{2y}}e^y=-1/e^y$

By now, I don't know how this derivative can be interpreted.

Solution 2:

For the first one, the notation of partial differentials is ambiguous enough to not know what you are asking for. however, I can calculate the total differential for you, however it uses a different-than-usual notation (see my paper, "Extending the Algebraic Manipulability of Differentials"). In this notation, the "normal" second derivative is $\frac{d^2y}{dx^2} - \frac{dy}{dx}\frac{d^2x}{dx^2}$. The notation for what you are wanting would be:

$$ \frac{d\left(\frac{dy}{dx}\right)}{dy} = \frac{\frac{d^2y}{dx} - \frac{dy\,d^2x}{dx^2}}{dy} = \frac{d^2y}{dx\,dy} - \frac{d^2x}{dx^2} $$

For the second problem, there's not enough information to actually calculate a solution. The integral is really $\int \frac{dy}{dx}\,dy$. You haven't separated variables, and this isn't a linear combination of functions, so it isn't the result of differentiating a single function of two variables.