# Variable transformation in a derivative

1. Jun 27, 2012

### Joschua_S

Hi

Maybe I don't see the wood because of all the trees, but:

You have a second derivative $\frac{\mathrm{d}^2}{\mathrm{d}x^2} e^{-ax} \cdot u(ax)$

Now you make the variable transformation $w=ax$

How to express

$\frac{\mathrm{d}^2}{\mathrm{d}w^2}$

Thanks
Greetings

2. Jun 27, 2012

### mathman

Your question is confusing. Are you substituting w = ax in the original function or after you have the second derivative in x?

3. Jun 27, 2012

### HallsofIvy

If you let w= ax, then $f(x)= e^{-ax}u(ax)$ becomes $f(w)= e^ww[/tex]. Of course, then, [itex]df/dw= e^ww+ e^w= e^w(w+1)$ and the $d^2f/dw^2= e^w(w+ 1)+ e^w= e^w(w+ 2)$.

Nothing unusual about that. Note, however, that while $df/dx= (df/dw)(dw/dx)$, it is NOT true that "$d^2f/dx^2= (d^2f/dw^2)(dw/dx)$. Rather, $d^2f/dx^2= d/dx(df/dx)= d/dx((df/dw)(dw/dx))= d/dx(df/dw)+ (df/dw)(d^2w/dx^2)$$= (d/dw(df/dw))(dw/dx)+ (df/dw)(d^2w/dx^2)=(d^2f/dw^2)(dw/dx)+ (df/dw)(d^2w/dx^2)$.