1. ## A 'trivial' question...

It is well known that the derivative of a 'function of function' $\displaystyle y\{u(x)\}$ is...

$\displaystyle \frac {dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}$

A question for you: does it exist a similar expression for the second derivative $\displaystyle \frac {d^{2} y}{dx^{2}}$?...

Any help will be greatly appreciated!...

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$

2. Originally Posted by chisigma
A question for you: does it exist a similar expression for the second derivative $\displaystyle \frac {d^{2} y}{dx^{2}}$?...
The answer is usually expressed in slightly different terms.
If $\displaystyle y=f(u)$ and $\displaystyle u=g(x)$ where $\displaystyle f~\&~g$ are both twice differentiable then
$\displaystyle \frac{{d^2 y}}{{dx^2 }} = \frac{{d^2 y}}{{du^2 }}\left( {\frac{{du}}{{dx}}} \right)^2 + \frac{{dy}}{{du}}\frac{{d^2 u}}{{dx^2 }}$.

3. Originally Posted by chisigma
It is well known that the derivative of a 'function of function' $\displaystyle y\{u(x)\}$ is...

$\displaystyle \frac {dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}$

A question for you: does it exist a similar expression for the second derivative $\displaystyle \frac {d^{2} y}{dx^{2}}$?...

Any help will be greatly appreciated!...

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$

$\displaystyle \left(\frac{d^2 y}{du^2}\cdot \frac{du}{dx}\right)\left(\frac{du}{dx}\right)+\le ft(\frac{d^2 u}{dx^2}\right)\left(\frac{dy}{du}\right)$

Edited:- I'm late , Plato beat me .

4. Originally Posted by Plato
The answer is usually expressed in slightly different terms.
If $\displaystyle y=f(u)$ and $\displaystyle u=g(x)$ where $\displaystyle f~\&~g$ are both twice differentiable then
$\displaystyle \frac{{d^2 y}}{{dx^2 }} = \frac{{d^2 y}}{{du^2 }}\left( {\frac{{du}}{{dx}}} \right)^2 + \frac{{dy}}{{du}}\frac{{d^2 u}}{{dx^2 }}$.
The reason is that $\displaystyle \frac{d^2y}{dx^2} = \frac d{dx}\Bigl(\frac{dy}{dx}\Bigr) = \frac d{dx}\Bigl(\frac{dy}{du}\frac{du}{dx}\Bigr) = \frac d{dx}\Bigl(\frac{dy}{du}\Bigr)\frac{du}{dx} + \frac{dy}{du}\frac{d^2u}{dx^2}$ (product rule). But

$\displaystyle \frac d{dx}\Bigl(\frac{dy}{du}\Bigr) = \frac d{du}\Bigl(\frac{dy}{du}\Bigr)\frac{du}{dx} = \frac{d^2y}{du^2}\frac{du}{dx}$.

5. For you all: thanks very much! ...

... the purpose of my question is related to the following differential equation...

$\displaystyle x^{2} \cdot y^{''} + x\cdot y^{'} + y = \sin (\ln x)$ (1)

... that has been proposed in another section...

A 'brute force approach' to (1) seems to have little chances but if we operate the change of variables $\displaystyle u= \ln x$ , thanks to the formula you have supplied, we have...

$\displaystyle \frac {dy}{dx} = \frac{dy}{du}\cdot \frac{du}{dx} = \frac{1}{x}\cdot \frac{dy}{du}$

$\displaystyle \frac{d^{2}y}{dx^{2}} = \frac{d^{2}y}{du^{2}}\cdot (\frac{du}{dx})^{2} + \frac{dy}{du}\cdot \frac{d^{2}u}{dx^{2}} = \frac{1}{x^{2}}\cdot (\frac{d^{2}y}{du^{2}} - \frac{dy}{du})$ (2)

... and the (1) becomes...

$\displaystyle y^{''} + y = \sin u$ (3)

... that is much more tractable [linear DE with constant coefficients...] and has as solution...

$\displaystyle y = c_{1} \cos u + c_{2} \sin u + \frac{u}{2}\cdot \cos u$ (4)

thanks again! ...

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$