I am given:

$$\tfrac{1}{4}(x+y)^2 + \tfrac{1}{9}(x - y)^2 = 1$$

Using the chain rule, and factoring out $y'$, I'm left with:

$$y' \left(\tfrac{1}{2}(x+y) - \tfrac{2}{9}(x-y)\right) = 0$$

Now I need to isolate $y'$ but I'm not sure how.

Should I do:

$$y' = \frac{1}{\tfrac{1}{2}(x+y) - \tfrac{2}{9}(x-y)}$$

Am I going about this question the correct way?



In the implicit differentiation of an equation $f(x,y(x))=C$ you also need to compute the derivative for the x variable, i.e.,


The $f_x$ part is missing in your derivative.

  • $\begingroup$ Could you show me in terms of my expression? I'm not sure I understand... the derivitive of x is 1 so with chain rule: (x+y)^2 -> 2(x+y) * (1 + y') oh, I think that's my issue isn't it... $\endgroup$ – jmasterx Feb 20 '14 at 13:07
  • $\begingroup$ Yes, exactly that. Now collecting those expressions should lead to the correct result. $\endgroup$ – Dr. Lutz Lehmann Feb 20 '14 at 13:28

Your first step is wrong. It should be $$\frac{1}{2}(1+y')(x+y)+\frac{2}{9}(1-y')(x-y)=0$$

  • $\begingroup$ Thanks, I just realized that :P $\endgroup$ – jmasterx Feb 20 '14 at 13:11

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