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Is d(x^2)/x^2 = dx/x ?

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(edited 9 years ago)
what does the d represent? Is it a function, or are you differentiating and have missed off differentials on the bottom?
Original post by Vapor
does the d prevent one from just taking the square root of top and bottom ?


Your notation is meaningless. Can you clarify what you actually mean? I'm fairly sure the answer will be yes by the way.
Even without any d's, I would advise against taking the root of the top and bottom of a fraction :smile:
Original post by Desk-Lamp
Even without any d's, I would advise against taking the root of the top and bottom of a fraction :smile:


Hahahahahaha
Reply 5
Avatar for davros
davros
16
Original post by Vapor
does the d prevent one from just taking the square root of top and bottom ?


Why would you want to take the square root?

If d and x are just numbers you can use normal algebra to simplify
Original post by davros
Why would you want to take the square root?

If d and x are just numbers you can use normal algebra to simplify


I think take the square root was supposed to mean divide by x in this case. But, as usual, the OP will probably never reappear to enlighten us.
It's a case where it looks like a misunderstanding in basic differentiation, but could actually be an advanced topic which we're missing completely.
Original post by Desk-Lamp
It's a case where it looks like a misunderstanding in basic differentiation, but could actually be an advanced topic which we're missing completely.


Probably yes to the former but no to the latter.
How about this for a solution:

lety=x2 let y = x^2

dydx=2x \frac{dy}{dx} = 2x

dy=2xdx dy = 2x dx

dyx2=2dxx \frac{dy}{x^2} = 2\frac{dx}{x}

d(x2)x2=2dxx \frac{d(x^2)}{x^2} = 2\frac{dx}{x}

Following it through like this it almost makes sense, but the seperated differentials make me shudder. And you certainly can't root the top/bottom of a fraction and expect the same value afterwards.
(edited 9 years ago)
Original post by Desk-Lamp
How about this for a solution:

lety=x2 let y = x^2

dydx=2x \frac{dy}{dx} = 2x

dy=2xdx dy = 2x dx

dyx2=2dxx \frac{dy}{x^2} = 2\frac{dx}{x}

d(x2)x2=2dxx2 \frac{d(x^2)}{x^2} = 2\frac{dx}{x^2}

Following it through like this it almost makes sense, but the seperated differentials make me shudder. And you certainly can't root the top/bottom of a fraction and expect the same value afterwards.


Original post by Mr M
Probably yes to the former but no to the latter.


Original post by davros
Why would you want to take the square root?

If d and x are just numbers you can use normal algebra to simplify


Original post by Omghacklol
Hahahahahaha




Posted from TSR Mobile

Original post by Vapor
does the d prevent one from just taking the square root of top and bottom ?


Dont know what everyone in this thread is on about on a simple question as this

The answer is simply yes, they both equal to 1, assuming the OP means d(x^2)/dx^2 and dx/dx

When y =x , dy/dx is 1, dy = dx , therefore dx/dx = 1

Where y = x^2 and u = y , du/dy =1 , therefore d(x^2)/dy = 1, and since du = dy
d(x^2)/x^2 is simply 1

I tried to make it to the easiest possible for you guys to understand, in reality this should be done in a second by sight, wait , actually it is common sense

100% in all further math modules master race checking in

Posted from TSR Mobile
(edited 9 years ago)
Original post by Med_medine
Posted from TSR Mobile

Dont know what everyone in this thread is on about on a simple question as this

The answer is simply yes, they both equal to 1, assuming the OP means d(x^2)/dx^2 and dx/dx

When y =x , dy/dx is 1, dy = dx , therefore dx/dx = 1

Where y = x^2 and u = y , du/dy =1 , therefore d(x^2)/dy = 1, and since du = dy
d(x^2)/x^2 is simply 1

I tried to make it to the easiest possible for you guys to understand, in reality this should be done in a second max by sight

100% in all further math modules master race checking in

Posted from TSR Mobile


This could be it, but I'm not convinced there's actually a typo. No way to tell until the poster comes back :smile:
Original post by Desk-Lamp
This could be it, but I'm not convinced there's actually a typo. No way to tell until the poster comes back :smile:


Or he did it deliberately for us to fire our thoughts in all directions

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Avatar for davros
davros
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Original post by Med_medine
Posted from TSR Mobile



Dont know what everyone in this thread is on about on a simple question as this

The answer is simply yes, they both equal to 1, assuming the OP means d(x^2)/dx^2 and dx/dx



The answer could be just about anything depending on whatever assumptions you make. We're trying to establish exactly what it is that the OP does mean :smile:
Original post by Med_medine
Posted from TSR Mobile



Dont know what everyone in this thread is on about on a simple question as this

The answer is simply yes, they both equal to 1, assuming the OP means d(x^2)/dx^2 and dx/dx

When y =x , dy/dx is 1, dy = dx , therefore dx/dx = 1

Where y = x^2 and u = y , du/dy =1 , therefore d(x^2)/dy = 1, and since du = dy
d(x^2)/x^2 is simply 1

I tried to make it to the easiest possible for you guys to understand, in reality this should be done in a second by sight, wait , actually it is common sense

100% in all further math modules master race checking in

Posted from TSR Mobile


'master race' lolol, fair play
Original post by Omghacklol
'master race' lolol, fair play


I dont get what you are trying to say

Posted from TSR Mobile
Having seen a previous post by the poster, I am fairly sure this is actually a problem along the lines of:

Let V = x^2. Then dV=2xdxdV = 2x dx and so dividing through by V = x^2 we get dVV=2dxx\dfrac{dV}{V} = 2\dfrac{dx}{x} . If you get now rewrite V as x^2 you end up with d(x2)x2=2dxx\dfrac{d(x^2)}{x^2} = 2\dfrac{dx}{x}.

Note the 2, which means what he wants to be true is not true.

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