does y always get differentiated to dy/dx?Watch

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#1
Hey,

Just came across a few questions where I had to differentiate 'substitution v=y^-3'
for differential equations in terms of dy/dx, x, and y.

I thought that y would always differentiate into dy/dx hence differentiating v would give me an anwer in terms of dy/dx.

But I found out that that's not always the case?

When exactly do you differentiate y into dy/dx and when do you know y is just some constant so you differentiate as normal (eg. -3y^-4)?
0
2 years ago
#2
It depends on what letter the term you're differentiating actually is.
For example, let's say y = x^2
then, dy/dx = 2x

But, if y = t^2
then, dy/dt = 2t

Also, lets say we replaced y with g. g = x^2
then, dg/dx = 2x
and if we do p = i^2
then dp/di = 2i

In your example (v = y^-3), differentiating v would give you an answer in terms of dv/dy.

Hope that helped .
1
2 years ago
#3
it depends on what you are looking for , if it was dv/dy then you would just differentiate y as you would usually do with x but if it was dy/dx then you would have to do it implicitly as you are differentiating with respect to x. so in general you would differentiate with respect to the term on the bottom e. if dy/dx then respect to x or dv/dy then respect to y and so on. and any term that isn't the same variable as the one you are differentiating with respect to you would do implicitly.
1
2 years ago
#4
(Original post by liemluji)
Hey,

Just came across a few questions where I had to differentiate 'substitution v=y^-3'
for differential equations in terms of dy/dx, x, and y.

I thought that y would always differentiate into dy/dx hence differentiating v would give me an anwer in terms of dy/dx.

But I found out that that's not always the case?

When exactly do you differentiate y into dy/dx and when do you know y is just some constant so you differentiate as normal (eg. -3y^-4)?
Think of it in terms of implicit differentiation, which is always done.

Differentiating y wrt x is 1 then you multiply by So if v=y^-3 and you diff wrt y, you would write But if you want to diff. wrt x and there are no x terms, you would simply use the chain rule.
1
#5
(Original post by dididid)
it depends on what you are looking for , if it was dv/dy then you would just differentiate y as you would usually do with x but if it was dy/dx then you would have to do it implicitly as you are differentiating with respect to x. so in general you would differentiate with respect to the term on the bottom e. if dy/dx then respect to x or dv/dy then respect to y and so on. and any term that isn't the same variable as the one you are differentiating with respect to you would do implicitly.
So does 'differentiating implicitly' mean differentiate y as dy/dx?
and for my example that I wrote in the beginning, is it possible to even differentiate v with respect to x? If so, what would y be differentiated as?
0
2 years ago
#6
(Original post by liemluji)
So does 'differentiating implicitly' mean differentiate y as dy/dx?
and for my example that I wrote in the beginning, is it possible to even differentiate v with respect to x? If so, what would y be differentiated as?
if you're differentiating with respect to x then yes, d/dx(y) = dy/dx.
v differentiated with respect to x would be similar i.e. dv/dx.

To get dy/dx where you haven't been given y explicitly but you are told some variables are functions of y you need to manipulate the product rule. So let's say you have v=x^2 and you want dy/dx

dy/dx = dy/dv x dv/dx

simple enough right?

So dv/dx = 2x

and it follows that dy/dx = (dy/dv)2x

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1
#7
(Original post by Zxphyrs)
if you're differentiating with respect to x then yes, d/dx(y) = dy/dx.
v differentiated with respect to x would be similar i.e. dv/dx.

To get dy/dx where you haven't been given y explicitly but you are told some variables are functions of y you need to manipulate the product rule. So let's say you have v=x^2 and you want dy/dx

dy/dx = dy/dv x dv/dx

simple enough right?

So dv/dx = 2x

and it follows that dy/dx = (dy/dv)2x

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Thank you so much! That was clear enough for me to understand 0
2 years ago
#8
Haha it's a pleasure

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