1. Aug 17, 2016

What is $\frac{d}{dx}(\frac{x}{y^2})$?

Please tell me is it correct or not:

$\frac{d}{dx}(\frac{x}{y^2}) = \frac{[\frac{d}{dx}(x)] ⋅ (y^2) - (x) ⋅ [\frac{d}{dx} (y^2)]}{(y^2)^2}$
$= \frac{(x) ⋅ (y^2) - (x) ⋅ (\frac{d}{dy} (y^2)) ⋅ \frac{dy}{dx}}{y^4}$
$= \frac{xy^2 - (x)(2y)(\frac{dy}{dx})}{y^4}$
$= \frac{xy^2 - 2xy(\frac{dy}{dx})}{y^4}$

2. Aug 17, 2016

### Staff: Mentor

Do you know what $\frac{d}{dx}x$ is? And isn't there another $y$ that could be canceled?

3. Aug 17, 2016

### Mastermind01

I think you made a typo - what is $\frac{d}{dx}(x)$ ?

P.S - fresh_42 posted at the time I was typing. Yes a y can be cancelled.

4. Aug 17, 2016

Yes, sorry, I've made a typo:

$\frac{d}{dx}(\frac{x}{y^2}) = \frac{[\frac{d}{dx}(x)] ⋅ (y^2) - (x) ⋅ [\frac{d}{dx} (y^2)]}{(y^2)^2}$
$= \frac{(1) ⋅ (y^2) - (x) ⋅ (\frac{d}{dy} (y^2)) ⋅ \frac{dy}{dx}}{y^4}$
$= \frac{y^2 - (x)(2y)(\frac{dy}{dx})}{y^4}$
$= \frac{y^2 - 2xy(\frac{dy}{dx})}{y^4}$

What y could be cancelled?

5. Aug 17, 2016

### Staff: Mentor

You differentiated something with $y^{-2}$ which would give you something with $y^{-3}$ but your result is something with $y^{-4}$.

6. Aug 17, 2016

Do you mean the $y^4$ in the denominator?

What's wrong with that?

The quotient rule is $\frac{u'v - uv'}{v^2}$ where $v = y^2$, am I right?

7. Aug 17, 2016

### Staff: Mentor

Nothing. If you like, you may even write $y^{100}$. Just said it's not the usual way. And you don't have to shout here.

8. Aug 17, 2016

I am not shouting. So, what do you mean with another $y$ that could be cancelled?

9. Aug 17, 2016

### Staff: Mentor

You have a common factor $y$ in each summand of the nominator and four of them in the denominator.
Or if you like to proceed with formulas: $ab+ac=a(b+c)$ and $\frac{y(y-2xy')}{y^4}=\frac{y-2xy'}{y^3}$.

10. Aug 17, 2016

OK, but is it correct of what I've done so far?

11. Aug 17, 2016

### Staff: Mentor

Yes.
And in case you want to differentiate it a second time, it will be a lot easier if $y$ is canceled out before. This also reduces the risk of making mistakes. Since $y$ must not be zero anyway, it may be canceled.

12. Sep 16, 2016

### rahul_26

There is a mistake in yours 2 step.
$$\because\;\dfrac{d}{dx}(x)=1\;\&\;\dfrac{d}{dx}(y)^2=2y\cdot\dfrac{dy}{dx}$$
So, your 2 step will be as-
$$\dfrac{d}{dx}\left(\dfrac{x}{y^2}\right)=\dfrac{1\cdot y^2-x\cdot2y\dfrac{dy}{dx}}{y^4}$$
$$=\dfrac{y^2-2xy\left(\dfrac{dy}{dx}\right)}{y^4}$$