So here is the question:

$$ \tan^{-1}\left(\frac{2x}{y}\right)=\frac{\pi x}{y^2} $$

When I solved it implicitly I got (with much pain in formatting it on this site :P):

$$ y^2=\pi \left(\frac{y^2-2xy\cdot y'}{y^4}\right)\cdot \left ( \sec^2\left(\frac{2x}{y}\right)\cdot (2y-2xy') \right ) $$

Now I know this sounds stupid but I don't know how to factor out y' because apparently I have derived correctly to the best of my knowledge and yet when I input (1,2) in my function and then check Wolphram Alpha, I get two different results (that shouldn't be the case)....

I'm at a loss as to what to do... Any help would be appreciated

  • $\begingroup$ +1 for the courtesy of writing your equations all nice and purdy ;) $\endgroup$ – David H Oct 18 '13 at 20:21
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    $\begingroup$ Did you start by rearranging to give $$\frac{2x}{y}=tan\left(\frac{\pi x}{y^2}\right)?$$ $\endgroup$ – George Tomlinson Oct 18 '13 at 20:34
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    $\begingroup$ no, I changed tan^-1 into 1/tan... Now that you mentioned it, I'm gonna try doing that... Thanks $\endgroup$ – StrugglingCalcStudent Oct 18 '13 at 20:40
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    $\begingroup$ So now my function after implicitly deriving it, is: $$ \frac{2y-2x\cdot y'}{y^2}=sec^2(\frac{\pi x}{y^2})\cdot \pi (\frac{y^2-2xy\cdot y'}{y^4}) $$ $\endgroup$ – StrugglingCalcStudent Oct 18 '13 at 20:43
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    $\begingroup$ I did it! I got it thanks to that simple method of solving it! Thank you @bluesh34 $\endgroup$ – StrugglingCalcStudent Oct 18 '13 at 20:51

Write your problem as F[x,y]=0 and perform total differentiation.

So 0 = dF = (dF/dx) dx + (dF/dy) dy and then y' = dy/dx = - (dF/dx) / (dF/dy)

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