How does dx/dy relate to dy/dx ?

uxa595

^
.

1/(dy/dx) = dx/dy

They are reciprocals.

Reply 3

watchthis

dy/dx is 1/dx/dy. is that what you meant?

Reply 4

nuodai

As the others have said, provided that

\dfrac{dy}{dx} \ne 0

. Whenever

\dfrac{dy}{dx} = 0

\dfrac{dx}{dy}

isn't defined. And likewise, whenever

\dfrac{dx}{dy}=0

\dfrac{dy}{dx}

isn't defined.

Reply 5

abbii

one is the reciprocal of the other
ie dx/dy = 1/ (dy/dx)

for example
if y=ax^2 + bx +c
then differentiating with respect to x gives
dy/dx = 2ax + b
but differentiating with respect to y gives
1=2ax(dx/dy)+b(dx/dy)
dx/dy = 1/(2ax+b)

(edited 13 years ago)

Reply 6

Crazy Paving

Original post by Mr M

They are reciprocals.

Indeed.

They're just each other flipped over, if that makes more sense, OP. :dontknow:

Reply 7

Mr M

Study Forum Helper

Original post by abbii

one is the inverse of the other

Are you sure about this?

Reply 8

abbii

Original post by Mr M

Are you sure about this?

i meant it in the sence that they are recipicols
i was wrong thinking about it
the inverse to dy/dx would be the integral of y with respect to x
i will edit earlier post

Reply 9

uxa595

so is it dy/dx = 1/(dx/dy)
or
1/(dy/dx)=dx/dy

Reply 10

Mr M

Study Forum Helper

Original post by uxa595

so is it dy/dx = 1/(dx/dy)
or
1/(dy/dx)=dx/dy

both

Reply 11

generalebriety

Original post by uxa595

so is it dy/dx = 1/(dx/dy)
or
1/(dy/dx)=dx/dy

What's the difference?

Reply 12

nuodai

Original post by abbii

i meant it in the sence that they are recipicols
i was wrong thinking about it
the inverse to dy/dx would be the integral of y with respect to x
i will edit earlier post

Strictly speaking that's not true, although it is for A-level purposes, so to avoid confusion I won't say any more. (But it's good to know that when you say that you're doing a bad, bad, evil, terrible thing).

Original post by uxa595

so is it dy/dx = 1/(dx/dy)
or
1/(dy/dx)=dx/dy

Both, because

a = \dfrac{1}{b} \iff b = \dfrac{1}{a}

Reply 13

generalebriety

Original post by nuodai

I don't think you'd be confusing anyone at A-level particularly much by giving a brief outline of the problem. Pathological cases aside, the problem is simply that, if you take a function f, integrate it, then differentiate the result, you get f back; however, if you take a function g, differentiate it, then integrate the result, you get g + (some constant) - so, in fact, any function that has an integral has other integrals too by just tweaking the constant. They're essentially the same, but not quite. So integration and differentiation are almost inverses in those cases, but not quite.

If anyone wonders what I mean by "pathological cases", well, for example... some functions just can't be differentiated everywhere (e.g. |x|), and likewise some functions can't be integrated everywhere (though coming up with an example of those requires a bit more creativity - imagine a function that fluctuates between positive and negative so erratically that its integral can't possibly converge), so it doesn't make sense to start comparing the derivative of the integral and "the" integral of the derivative.