# Using implicit differentiation to solve a function and stuck at factoring out y'.

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

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