1. ## Implicit Differentiaton

Could somebody please clarify for me what does it mean exactly, in simple terms?

Then i have the following problem i have some questions about:

$y^5 + 3x^2 = sin x - 4y^3$

Then we differentiate...

$(5y^4 . y') + (6x) = (cos x) - (12y^2 . y')$

Ok, now where did that y' come from and what exactly does it mean?

2. Originally Posted by janvdl

Could somebody please clarify for me what does it mean exactly, in simple terms?

Then i have the following problem i have some questions about:

$y^5 + 3x^2 = sin x - 4y^3$

Then we differentiate...

$(5y^4 . y') + (6x) = (cos x) - (12y^2 . y')$

Ok, now where did that y' come from and what exactly does it mean?
the y' is actually short for dy/dx

i think i told you what this notation meant before, dy/dx means "the derivative of y with respect to x"

in implicit differentiation we have to take account of what variable we are differentiating, and what variable we are differentiating with respect to. in the question you posted, you wanted to take the derivative with respect to x. so when we take the derivative of an x term, we attach dx/dx (we took "the derivative of x with respect to x"), however, since derivative notations can function as fractions, the dx's cancel and it becomes 1, so you don't see it, since it's like multiplying by one. however, when we take the derivative of a y term, we attach dy/dx (since we took "the derivative of y with respect to x"), and that doesn't cancel, so it stays. we write y' when it's understood what we are differentiating with respect to.

we need implicit differentiation to differentiate when we have mixed terms that we can't really separate. it would be hard to solve for y and then find y' that way, so we do it implicitly

3. Originally Posted by Jhevon
the y' is actually short for dy/dx

i think i told you what this notation meant before, dy/dx means "the derivative of y with respect to x"

in implicit differentiation we have to take account of what variable we are differentiating, and what variable we are differentiating with respect to. in the question you posted, you wanted to take the derivative with respect to x. so when we take the derivative of an x term, we attach dx/dx (we took "the derivative of x with respect to x"), however, since derivative notations can function as fractions, the dx's cancel and it becomes 1, so you don't see it, since it's like multiplying by one. however, when we take the derivative of a y term, we attach dy/dx (since we took "the derivative of y with respect to x"), and that doesn't cancel, so it stays. we write y' when it's understood what we are differentiating with respect to.

we need implicit differentiation to differentiate when we have mixed terms that we can't really separate. it would be hard to solve for y and then find y' that way, so we do it implicitly
Ok, so every y in the term has to be multiplied with y' ?

By the way Jhevon, u gota be online on msn before i can send you a message. Im using my phone for msn.

4. Originally Posted by janvdl
Ok, so every y in the term has to be multiplied with y' ?

By the way Jhevon, u gota be online on msn before i can send you a message. Im using my phone for msn.
yes. and if you were differentiating with respect to another variable besides x (which you will do when you start related rates) you would have to add x' to all the x terms as well

5. Originally Posted by Jhevon
which you will do when you start related rates...
Thats next years worries Jhevon.
Thanks for the help guys.

6. Originally Posted by janvdl
Thats next years worries Jhevon.
Thanks for the help guys.
actually, related rates is usually the topic taught right after (or very soon after) implicit differentiation, so it's closer than you think

7. Originally Posted by Jhevon
actually, related rates is usually the topic taught right after (or very soon after) implicit differentiation, so it's closer than you think
I'm actually not supposed to be doing this kind of stuff already.