0

I was solving some physics but stuck in some math I need to change variable x to y in $${d^2\psi(x) \over dx^2}$$ where $y=\alpha x$

I reached till $${d \over dx}({1\over \alpha }{d\psi \over dy})$$ Now should I use formula of ${u\over v}$ in differentiation but then I have to introduce another variable t then $${1\over \alpha}{d\over dx}\bigg({{d\psi \over dt}\over{dy \over dt}}\bigg)$$ I am very confused with another variable and unable to get feel.

Also I suppose writing $${d^2\psi(x) \over dx^2}={{d^2\psi(x) \over dt^2} \over{d^2(x) \over dt^2}}$$ is wrong

RKK
  • 408

2 Answers2

1

\begin{align} \frac{d^2\psi(x)}{dx^2} & = \frac{d}{dx}\left(\frac{d\psi(x)}{dx}\right) \\ & = \frac{dy}{dy}\cdot\frac{d}{dx}\left(\frac{d\psi(x)}{dx}\cdot\frac{dy}{dy}\right) \\ & = \frac{dy}{dx}\cdot\frac{d}{dy}\left(\frac{d\psi(x)}{dy}\cdot\frac{dy}{dx}\right) \\ & = \alpha^2\cdot\frac{d}{dy}\left(\frac{d\psi(x)}{dy}\right) \\ \end{align}

RiverX15
  • 888
1

Assuming $\alpha$ does not depend on $x$ for all derivation operators you have: $$\frac{d}{dx}= \frac{d}{d(y/\alpha)}=\alpha\frac{d}{dy}$$ so $$\frac{d^2\Psi(x)}{dx^2}=\alpha^2\frac{d^2\Psi(x)}{dy^2}=\alpha^2\frac{d^2\Psi(y/\alpha)}{dy^2}$$ You can go full circle by doing the chain rule here of course: $$\alpha^2\frac{d^2}{dy^2}\Psi(y/\alpha)=\alpha^2\frac{d}{dy}\Psi'(y/\alpha)\frac{1}{\alpha}=\alpha^2\Psi''(y/\alpha)\frac{1}{\alpha^2}=\Psi''(x)$$

mr_tuna
  • 111