[JeremyB]

I'm differentiating this function:



So far I've got this:



I need to get an expression for y' but I'm really struggling to separate them out. Could someone show me how to do this. I'm sure it's simple.

Thanks!

[aurao2003]

Have you tried cross multiplying? It should be possible to collect like terms for dy/dx and then use subject formula.

[Swayum]

If I asked you to solve this equation for a, could you do it?



Your question is no different.

[JeremyB]

(Original post by aurao2003)
Have you tried cross multiplying? It should be possible to collect like terms for dy/dx and then use subject formula.

Thanks for the reply. Do you mean multiplying out the and then collecting the y' terms? I thought about doing that, but is there another way that would be less time consuming in the exam?

[JeremyB]

(Original post by Swayum)
If I asked you to solve this equation for a, could you do it?



Your question is no different.

Thanks. So are you saying that multiplying out the brakets is the only way in this case?

[Swayum]

(Original post by JeremyB)
Thanks for the reply. Do you mean multiplying out the and then collecting the y' terms? I thought about doing that, but is there another way that would be less time consuming in the exam?

Well, you didn't need to divide through by 1-y' in the first place, right? So that saves you a couple of lines of working out, but no there's nothing shorter than rearranging the equation for y'.

[Swayum]

(Original post by JeremyB)
Thanks for the reply. Do you mean multiplying out the and then collecting the y' terms? I thought about doing that, but is there another way that would be less time consuming in the exam?

Also, just to clarify, multiplying out the above just means getting:



You obviously wouldn't expand out (x-y)^3.

[JeremyB]

(Original post by Swayum)
Also, just to clarify, multiplying out the above just means getting:



You obviously wouldn't expand out (x-y)^3.

Ah, thanks for that. I was thinking I had to multiply out the (x-y)^3.

So would this give me:

?

[Swayum]

(Original post by JeremyB)
Ah, thanks for that. I was thinking I had to multiply out the (x-y)^3.

So would this give me:

?

Nope, the numerator should be 4(x-y)^3 - 1 I think (although I could be wrong).

[JeremyB]

(Original post by Swayum)
Nope, the numerator should be 4(x-y)^3 - 1 I think (although I could be wrong).

Thanks a lot for your help.