relationship between dy/dx and dx/dy

If a car drive 50 miles per hour     dx/dt

How long does it take to drive 1 mile    dt/dx     ==> 1/50 hours

diplomat said:

Suppose that y=f(x) implicitly defines y as a function of x. differentiate both sides with respect to y:

How can you differentiate both sides with respect to y? My understanding of differentation is that, for y = f(x) it tells you how how  y changes with respect to x

How can you differentiate the function with respect to y? Perhaps this is obvious but I've never seen differentiation used this way. I've certainly never seen you differentiate y with respect to y.

Could you give me an example from first princples (i.e. f(x+dx) - f(x)/dx) type of approach) for a really simple equation like y = 2x^2 how you differentiate x wrt y and y wrt y. I'm only familiar with the gradient being (change in y /change in x). In other words i'm only familar with concepts like 'at this point, y is increasing twice as fast as x. I don't know how to conceptualise it if you differentiate wrt y

Many thanks

Hi infinity

You're great as ever.

I can see intuitively that dx/dy is the reciprocal of dy/dx. I just like to be able to prove things where possible.

As you said, i could take 2 reciprocal functions and look at the relationship between their derivative, but is there some generic formula that shows that dy/dx= 1/(dx/dy).

I think this might be what the diplomat is saying but I don;t understand the answer, because of my limiitations

You are always differentiating both sides its just not always said out loud. Per your example, you are actually taking the derivative of both sides with respect to x or multiplying by d/dx.

y = 2x^2  

Now take the derivative of both sides with respect to x. That is like saying multiply both sides by              d/dx.

(d/dx)*y=(2x^2)*d/dx

dy/dx= 4x or y'=4x

in the example initially showed you we were taking the derivative with respect to y so we started with
y=f(x)  now we take the derivative of both sides with respect to y (multiply both by d/dy)

(d/dy)*y=f(x)*(d/dy)

dy/dy= d(f(x))/dy
The derivative of anything with respect to itself is 1 so that is why (d/dy)*y=1 and then you follow the rest of the steps from there.

I hope that clarifies it. The way it is worded is confusing. When you take the derivative of an equation you are always taking the derivative of both sides, because in math whatever you do to one side of the equation you must do to the other.

>> How can you differentiate the function with respect to y?

You can differentiate any function you want with respect to y. The result is not necessarily meaningful, but you can ...


>> I've certainly never seen you differentiate y with respect to y.

dy/dy = 1 just like dx/dx = 1. See one of your previous questions : http://www.experts-exchange.com/Other/Math_Science/Q_24769627.html

Hi Diplomat,

I have looked at the question that you have referred to and I can follow this question and see how you get the answer. However, for some reason, and I am struggling with the generic explanation.

I know you have already taken the time to try and explain it again for me in different words but I don't get it still :(

So, if you are willing, perhaps we need to take it one step at a time :)

I had never thought of taking the derivative as being something that applied the same operation to both sides of the equation so I will try and think of it from that point of view

So taking the y = x^2 example. You said

"Take the derivative of both sides wrt x. That is like multiplying both sides by d/dx"

What exactly here is d/dx? I just thought it was notation but perhaps it means something else as I wasn't aware you could multiply something by it. Do you see what I mean? In my understanding, dy/dx can be included in an equation because it results in an expression in the same way that ln(x) results in an expression. To me, dy/dx ultimately becomes a number. I don't know what d/dx is. Perhaps i am thinking of it in completely the wrong terms.

thanks

I understand your confusion. I remember the same feeling as well. My line where I say " That is like multiplying both sides by d/dx" was just to try to clarify the point. There is no actual multiplication by a d/dx. You are applying it to both sides would have been better wording. Sorry for that miscommunication.
these are some points which hopefully clarify a little step by step.

1) Whatever is done to 1 side of an equation must be done to the other to keep both sides equal
2) the term dy/dx=y'. do you agree with that.
3) the term d/dx just means "take the derivative with respect to x".
4) so then dy/dx means "take the derivative of y with respect to x".
5) y = 2x^2, so if you are taking the derivative of y you are taking the derivative of 2x^2 (since they are the same (equal0
6) dy/dx = d(2x^2)/dx by subsitution.
7) d(2x^2)/dx means "take the derivative of 2x^2 with respect to x" which we know is 4x (or you can say 4x is the derivative of 2x^2 with respect to x).
8)so dy/dx = d(2x^2)/dx = 4x

When i am saying with respect to (variable), all that means is that that is the variable you are using to take the derivative. that terminology tends to come more into play in later calc when you have equations with multiple variables (i.e. f(x,y)= 4x^3+3xy-5y^2) that is when it is necessary to know what variable you are taking the derivative with respect to.

Hello

Lets take y = 2x^2

I understand that what must be done to one side of the equation must be done to the other to keep everything equal. However i don't see this line

dy/dx = 4x as being a case of applying the same operation to both sides

I fully understand that the derivative of y wrt to x is the same as the derivative of 2x^2 wrt x as y = 2x^2. I can see that d(2x^2)/dx is 4x because i know how to apply the differentation rules to the terms in x. so then i can see that dy/dx  = 4x because d(2x^2)/dx = 4x. So, i see how to get from 2x^2 to 4, but not directly from y to 4x.

So when i read dy/dx = 4, in my mind i am not reading 'take the derivate of both sides' . i do not read that i have applied the same operation to both sides. I have applied the operation to the RHS and shown the answer, but left the symbol for the operation on the LHS. Subtle difference maybe.

Again when i read something like

y = f(x)
dy/dy = d(fx)/dy

My brain does not say, 'take the derivative of both sides wrt to y'
My brain says, the derivative of y wrt y is the same as the derivative of fx wrt y as y and f(x) are the same value.

Hopefully, my way of looking at it is equally valid. It might be subtle difference to you but it sits better with me

>> So when i read dy/dx = 4, in my mind i am not reading 'take the derivate of both sides' . i do not read that i have applied the same operation to both sides.

If it helps, you can write out the extra step explicitly :

        y = 2x^2
        dy/dx = d(2x^2)/dx
        dy/dx = 4x

That way, you see clearly that both sides were derived with regard to x. The last step was just simplifying it (working it out) a bit further ...


>> My brain does not say, 'take the derivative of both sides wrt to y'
>> My brain says, the derivative of y wrt y is the same as the derivative of fx wrt y as y and f(x) are the same value.

Sure it is ... Because both statements mean the same thing.

These are essentially equivalent statements as far as I am concerned.

My brain does not say, 'take the derivative of both sides wrt to y'
My brain says, the derivative of y wrt y is the same as the derivative of fx wrt y as y and f(x) are the same value.

As they always say, everyone learns differently. From my point of view taking the derivative with respect to both sides makes more sense to me then how you are stating it, but they do mean the same thing. I think you now see the logic in the steps from y=2x^2 to dy/dx=4x.

So if you are still with me and haven't wanted to strangle me yet, that brings me back to the original question

I can see mathematically that dy/dx = 1/(dx/dy) because its just basic maths as pointed out by hathehariken

I can also differentiate a function and then differentiate its inverse and see that they are the same thing. What got this post so long was when the diplomat kindly gave a generic proof (that's probably not the right word, i know it has precise maths meaning) that I didn't understand.But i still don't get that because i don't fully get the chain rule. so i'll close his post and start another one about th chain rule!
So i'm going to close this question and probably ask another one about the chain rule.