# Chain Rule Question

1. Oct 25, 2010

### Juanriq

1. The problem statement, all variables and given/known data Reduce the order of a Cauchy-Euler Equation

2. Relevant equations $x = e^t \mbox{ and } \ln x = t$

3. The attempt at a solution
$\displaystyle \frac{d y}{d x} = \displaystyle \frac{d y}{d t} \displaystyle \frac{d t}{d x} = \displaystyle \frac{d y}{d t} \cdot \displaystyle \frac{1}{x}$
and thus
$\displaystyle \frac{d^2 y}{d x^2} = \displaystyle \frac{d y}{d t} \cdot \displaystyle \frac{-1}{x^2} + \displaystyle \frac{1}{x} \displaystyle \frac{d}{d x} \Bigl ( \displaystyle \frac{d y}{d t} \Bigl )$

Here is where I am getting stuck, specifically on $\displaystyle \frac{d}{d x} \Bigl ( \displaystyle \frac{d y}{d t} \Bigl )$ this step. I know what I should get...
$\displaystyle \frac{1}{x} \Bigl ( \displaystyle \frac{d^2 y}{d t^2} \cdot \displaystyle \frac{1}{x} \Bigl )$
But uhhh not getting it. Thanks in advance!

2. Oct 25, 2010

### vela

Staff Emeritus
You apply the chain rule again, except instead of applying it to y, you're applying it to dy/dt. It's a bit clearer if you look at it in terms of operators:

$$\frac{d}{dx} f = \frac{dt}{dx}\frac{d}{dt} f = \frac{1}{x}\frac{d}{dt} f$$

so

$$\frac{d}{dx} = \frac{1}{x} \frac{d}{dt}$$

3. Oct 25, 2010

### Juanriq

Ahhh...very nice! Thanks a bunch!