# does y always get differentiated to dy/dx?

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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)?

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)?

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#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 .

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 .

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#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.

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#4

(Original post by

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)?

**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)?

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.

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(Original post by

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.

**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.

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?

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#6

(Original post by

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?

**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?

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

Posted from TSR Mobile

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(Original post by

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

Posted from TSR Mobile

**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

Posted from TSR Mobile

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