we say at derivatives that d/dx is just an operator which applies to a function.but when we come to the differentials we assume dx and dy to be in a ratio.Could any one help me at this? And infact we ourselves divides "delta x" in the derivative process
and after taking limit ,i don't know how do we say its not a ratio?

Question

we say at derivatives that d/dx is just an operator which applies to a function.but when we come to the differentials we assume dx and dy to be in a ratio.Could any one help me at this? And infact we ourselves divides "delta x" in the derivative process       
and after taking limit ,i don't know how do we say its not a ratio?

anonymous · Answer

I think of the derivative operator as being one that takes a function and returns back a new function.  The derivative is still a ratio, one that is saying 'at what rate to y change for some change in x. For  $$y=x$$ the derivative is 1, or $$y/x=1/1$$For $$y=x ^{2}$$ $$y \prime=2x$$ or $$y/x=2/1$$. Maybe the confusion comes from the fact that we don't normally write the derivative as a ratio?

anonymous · Answer

No!! the derivative is not a ratio... dy/dx means that you have applied the differential operator d/dx to the function 'y'.......dy/dx  represents an operation.

While, the differentials dx and dy represent numbers, not an operator or a part of it.... dx and dy are the infinitesimally small changes in the function when limit is approached.....

That is how notation has been written. No new notation was used by our predecessors.. So that is the irony. While doing differential equations or evaluating an integral .. the dx and dy are differentials...

anonymous · Answer

So bhaweshwebmaster, are there actually two concepts here? First, the 'dy/dx' as an operator applied to a function to get its derivative, and second, the differentials dy and dx are placeholders for the very small changes in the limit? I get the feeling there is a deeper meaning to 'differentials' that I'm not getting.

anonymous · Answer

Yes, let me use an example...

let f(x) = y = x^2 be a function of x. then,
 
its derivative is given as dy/dx = f '(x) = 2x...   So, the derivative of the function at x=2 is  f'(2) =4

But, the differential of y will be  dy= 2x .dx ... where dy is the change in y as a result of the infinitesimal change in the value of x quantified by dx

So, the differential of y at x=2 can be calculated as:

Let us change the x from x=2 to x=2.001... 

then the change in y will be dy= 2*2* (2.001-2) = 0.004  [ I would still complain that this looks same as derivative.. this is where the inventors of calculus could have done a better job)

{we can check the dy by calculating the actual change manually. }

f(2) = 4 and f(2.0010 = 4.004001, so the change in y is 0.004001.. which is reasonably correct... 

the value of dy will coincide the exact change as we make dx infinitesimally small.. like 0.000000001....  more smaller the dx,  better is the answer from differntial ]

I hope this makes sense to you....

anonymous · Answer

Now, you can see the derivative of y at x=2 is 4, while its differential at the very same point is 0.004.  A drastic difference... isn't it?

anonymous · Answer

Make sense. The derivative gives us an exact value for the rate of change, whereas the differentials give an approximate value that converges on the exact value of f(x+dx)-f(x) in the limit.

anonymous · Answer

@ariyama where did you find this idea of derivative as a ration?

anonymous · Answer

i mean if we just look at the way derivative is being define it says
change in y  divided by change in x "under the limit delta x goes to zero"
now if we ignore the limit for a while its an average rate and its simply a ratio (which so obvious).
Now i dont know how can some one say under the limit delta x goes to zero the derivative is not a ratio.Infact the average rate of change from where the derivative is originated is definately is pure ratio?

anonymous · Answer

and in ma point of view at derivative delta x approaches to zero and we really really mean that.
but in differentials delta x shud be small to get a gud approximation for the change in y.
and wen delta x goes larger and larger so you are geting as far from where the tangency point.

anonymous · Answer

differentials is just a kind of technique to find change in y.and its impossible to get a result without any error b/c the slope of the curve doesnt remain the same at each point (so it better to take dx as small possible ).
if u apply differential at a striaght line no matter how LARGE dx is it gonna give you no error b/c striaght line has same slope at each point.
e.g
y=2x+3
let x1=10  and x2=20 so dx=10
change in y=y2-y1=(2(20)+3) - (2(10)+3)=20
now by diferential
dy =derivative of function at x1  times dx
derivative of function at x1=2 and dx=10

so dy=2x10=20 (no error at all)

anonymous · Answer

@jk_16, I hadn't. I was trying to answer Usman93's question. My answer was the only way I could think of  treating it as ratio.

anonymous · Answer

well watever happens it does just after taking the limit..
but how? newton knows bettr.. lolx

anonymous · Answer

@usman93 buddy.. your earlier proof with  y=2x+3 , that showed no error at all is okay for linear function. Try some bi-quadratic, cubical or functions having higher powers of the variables....
Then error will start surfacing... and that is when only a limit will give you exact answer.. There is a topic called 'application of derivatives' that may help in better understanding of this issue..

anonymous · Answer

yeah b/c the slope of a curve doesnt remain the same at each point unlike a straight line..