(ASK) Implicit Differentiation

[askor]

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}##

 

[fresh_42]

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}##

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

 

[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.

 

[askor]

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.

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?

 

[fresh_42]

What y could be cancelled?

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

 

[askor]

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

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

What's wrong with that?

Isn't I am already correct?

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

 

[fresh_42]

What's wrong with that?

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.

 

[askor]

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.

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

Please explain, I don't understand.

 

[fresh_42]

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}##.

 

[askor]

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}##.

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

 

[fresh_42]

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

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.

 

[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-
$$\dfrac{d}{dx}\left(\dfrac{x}{y^2}\right)=\dfrac{1\cdot y^2-x\cdot2y\dfrac{dy}{dx}}{y^4}$$
$$=\dfrac{y^2-2xy\left(\dfrac{dy}{dx}\right)}{y^4}$$
I think this will help you in learning implicit differentiation.
http://www.actucation.com/calculus-...ative/implicit-derivative-and-its-application

 

[askor]

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-
$$\dfrac{d}{dx}\left(\dfrac{x}{y^2}\right)=\dfrac{1\cdot y^2-x\cdot2y\dfrac{dy}{dx}}{y^4}$$
$$=\dfrac{y^2-2xy\left(\dfrac{dy}{dx}\right)}{y^4}$$
I think this will help you in learning implicit differentiation.
http://www.actucation.com/calculus-...ative/implicit-derivative-and-its-application

Yes you're right. Thank you for the correction.