I am trying to differentiate this equation.

enter image description here

I have done it one way and got this solution (I don't know how to put it into Math formatting):


I used the quotient rule, and took out $y$ as a constant.

However when I looked at the solutions, my lecturer got:


I've simplified her solution so I know mine is right. However in order to complete the rest of the question it's far easier to have it in her form.

Can someone please walk me through, step by step, how to get this solution? I'm getting myself very confused.

Thanks in advance.

Original Question: I am working on part a.3. I have to find the Jacobian, and set the trace equal to zero and to start I need to find the derivative of x(dot) with respect to x. enter image description here

  • $\begingroup$ I am puzzled by this. You say you want to differentiate and equation. The derivative of an equation is, again, an equation. But what both you and your teacher give are expressions, not equations. $\endgroup$ – user247327 Jan 4 at 13:15
  • $\begingroup$ Apologies, it's probably me not being clear on my understanding of the differences between those. Does an equation have an =? $\endgroup$ – MathsIsFun Jan 4 at 13:17
  • $\begingroup$ Please take the time to enter crucial parts of your question as text instead of pasting pictures. Your question should be comprehensible without images, which are neither searchable nor accessible to screen readers. See math.meta.stackexchange.com/a/10992/265466. $\endgroup$ – amd Jan 4 at 20:40

You've multiplied $x$ with each of the terms in the parenthesis, and you took the derivative. You got the first two terms right. The last one is $$-\frac{xy}{1+x}$$You can use product rule, where the first function is $x$ and the second is $-y/(1+x)$. So the derivative of that is $$-\frac y{1+x}-x\frac{-y}{(1+x)^2}$$

If you use the quotient rule, you don't have a constant $y$ at the numerator, but you have $xy$ instead.

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  • $\begingroup$ So if the second function is -y/(1+x) do I think use the quotient rule to differentiate this? Or just treat y as a constant and differentiate it that way? $\endgroup$ – MathsIsFun Jan 4 at 13:20
  • $\begingroup$ @Emily you already did it right, except that you forgot to multiply it by $-x$ $\endgroup$ – Andrei Jan 4 at 13:22
  • $\begingroup$ Yes I see now, just written it down properly. Thank you for answering my question, I was getting very confused! $\endgroup$ – MathsIsFun Jan 4 at 13:23

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