maths differentiation question help

Dazzling7dreams

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Can u guys please help me with this question?
Do we get partial fractions for each and then differentiate?

From a bit of playing, brute force is going to be a real pain, whether you use partial fractions or not.

You need to work insightfully with this to get it to come out elegantly.

To start, I'd put everything (bar the 1) over a common denominator. Looking at the numerator, think about roots of polynomials, and how you might use that. (Edit: Not fully worked it through in detail, so I could be wrong in this approach, but it looks promising.)

Where's the question from? It looks STEP like.

(edited 4 years ago)

Reply 3

Dazzling7dreams

Thank you for replying 😄one of my friends sent it to me saying its an AL question however I didn't come through such question before😑.

Reply 4

ghostwalker

Original post by Dazzling7dreams

Thank you for replying 😄one of my friends sent it to me saying its an AL question however I didn't come through such question before😑.

Roots of polynomials is Further Maths, so if you're not doing that, try comparing the numerator with what the denominator would look like if expanded - they're very close.

Reply 5

mqb2766

One way of doing it is to combine everything into a common fraction. Keep the denominator factorized, but expand / simplify the numerator. Things are much simpler.
Then take natual logs and differentiate.
Finally, put everything over the common denominator and simplify the numerator.
The factorized numerator/denominator can then be split up in the final step

Its about 1/2 -> 1 page of working. The trick to this way is to recognose that the answer is "*y" on the right hand side. So that could come from 1/y*dy/dx on the left, so natural logs is one way to go.

Reply 6

ghostwalker

Original post by mqb2766

"ln" - nice idea, PRSOM

Reply 7

Physics Enemy

Original post by ghostwalker

"ln" - nice idea, PRSOM

It was my idea first lol, I posted a near solution (before mqb's post) but it got removed. Whole thing reduces to y = x^3/(x - a)(x - b)(x - c), take logs and differentiate for (1/y)dy/dx = 3/x + 1/(a - x) + 1/(b - x) + 1/(c - x). Only issue was it didn't quite match OP's answer.

(edited 4 years ago)

Reply 8

ghostwalker

Original post by Physics Enemy

You seem to be in the habit of posting almost full solutions.

Reply 9

Physics Enemy

Original post by mqb2766

I got up to (1/y)y' = (1/x)[3 + x/(a - x) + x/(b - x) + x/(c - x)] this morning but couldn't simplify to OP's answer.

I just realised, let z1 = (a - x), z2 = (b - x), z3 = (c - x) and you easily get [a/z1 + b/z2 + c/z3] as required.

(edited 4 years ago)

Reply 10

mqb2766

Original post by Physics Enemy

Original post by ghostwalker

You seem to be in the habit of posting almost full solutions.

It may have been your idea, but I didn't see your post. I'm not in the habit of repeating other peoples contributions and I explicit;y try to avoid it. If I respond at the same time as another person, I usually drop out and let the other person work through the problem.

There are a reasonable number of threads where Ive replied with some hints to allow the OP to try and work through them, and before they reply, you've provided a near full solution. Doing the last calculation doesn't allow the OPs to make mistakes in setting up the problem, choosing the method, ... There is always a matter of style about how someone responds and we're not always 100% cobsistent, but kids should be allowed to make mistakes and think about why they made them. They' shoildn't be expecting a homework service,.

You're obviously capable and can offer some good advice. The sticky at the start the forum recommends that people offer hints not solutions and if someone responds to give them (and the OP) enough time to work through the problem.