# From dy/dx to d/dx

1. Aug 19, 2011

### htdIO

Hi all,

This is not strictly a DE question, but I came across this while working on one. This isn't the first time I got this and I just can't remember this from my 1st year maths. Some knowledge would be greatly appreciated. In the answer they do the following:

$(\frac{1}{x})(\frac{dy}{dx}) - (\frac{1}{x^2})y \Rightarrow (\frac{d}{dx})[(\frac{1}{x})y]$

Now I want to know how? I just cant simplify it. Silly question, but need the help!

Thanks

2. Aug 19, 2011

### I like Serena

Welcome to PF, htdIO!

Are you familiar with the chain rule?

It is: $\frac d {dx} f(y(x)) = \frac {df} {dy} \frac {dy} {dx}$

Do you know how to apply this?

3. Aug 19, 2011

### htdIO

Hi and thanks!

I do know it. Just not quite sure how I should be applying it here? I've scribbled quite a bit down here now, trying to combine this with the product rule. Or am I heading in the wrong direction?

4. Aug 19, 2011

### I like Serena

Sorry, you're right. You need to apply the product rule.
Do you know how to apply it to: $(\frac{d}{dx})[(\frac{1}{x})y(x)]$?

5. Aug 19, 2011

### htdIO

Haha, aah thanks. I must be more tired than I thought...
I'm guessing the only way to 'see' this (like they did it), is by recognizing it and a bit of practice?

6. Aug 19, 2011

### I like Serena

Hah, after all the practice I got, I thought you needed the chain rule!
So much for all that practice!

7. Aug 19, 2011

### htdIO

Halfway through I actually remembered the quotient rule, which should make it quicker ;) Anyway, thanks again for getting me on the right track!

8. Aug 19, 2011

### I like Serena

Neh, the quotient rule is not quicker in this case.
But good you remembered it!