2
$\begingroup$

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?

Thanks

$\endgroup$
1
$\begingroup$

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.,

$$0=f_x(x,y(x))+f_y(x,y(x))y'(x)$$

The $f_x$ part is missing in your derivative.

$\endgroup$
  • $\begingroup$ 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
1
$\begingroup$

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

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.