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Implicit differentiation (algebra question)

I'm differentiating this function:

(xy)4=x+y+5 (x-y)^4 = x + y + 5

So far I've got this:

4(xy)3=1+y1y 4(x-y)^3 = \dfrac{1+y'}{1-y'}

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. :smile:

Thanks!
(edited 11 years ago)
Reply 1
Have you tried cross multiplying? It should be possible to collect like terms for dy/dx and then use subject formula.
Reply 2
If I asked you to solve this equation for a, could you do it?

5=1+a1a\displaystyle 5 = \frac{1+a}{1-a}

Your question is no different.
Reply 3
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. :smile: Do you mean multiplying out the 4(xy)3(1y) 4(x-y)^3(1-y') 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?
Reply 4
Original post by Swayum
If I asked you to solve this equation for a, could you do it?

5=1+a1a\displaystyle 5 = \frac{1+a}{1-a}

Your question is no different.

Thanks. :smile: So are you saying that multiplying out the brakets is the only way in this case?
Reply 5
Original post by JeremyB
Thanks for the reply. :smile: Do you mean multiplying out the 4(xy)3(1y) 4(x-y)^3(1-y') 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'.
Reply 6
Original post by JeremyB
Thanks for the reply. :smile: Do you mean multiplying out the 4(xy)3(1y) 4(x-y)^3(1-y') 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:

4(xy)3(1y)=4(xy)34y(xy)3 4(x-y)^3(1-y') = 4(x-y)^3 - 4y'(x-y)^3

You obviously wouldn't expand out (x-y)^3.
Reply 7
Original post by Swayum
Also, just to clarify, multiplying out the above just means getting:

4(xy)3(1y)=4(xy)34y(xy)3 4(x-y)^3(1-y') = 4(x-y)^3 - 4y'(x-y)^3

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

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

So would this give me:

y=14(xy)34(xy)3+1y' = \dfrac{1-4(x-y)^3}{4(x-y)^3 + 1} ?
Reply 8
Original post by JeremyB
Ah, thanks for that. :smile: I was thinking I had to multiply out the (x-y)^3.

So would this give me:

y=14(xy)34(xy)3+1y' = \dfrac{1-4(x-y)^3}{4(x-y)^3 + 1} ?


Nope, the numerator should be 4(x-y)^3 - 1 I think (although I could be wrong).
Reply 9
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. :wink:

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