MHB Implicit differentiation difficult

[Fermat1]

Given that $y'=\frac{tan(y)}{1-xsec^2(y)}$, find y'' in terms of $x$ and $y$ only.

I've done this and checked my work several times but my answer does not agree with wolfram alpha. Sorry for not posting my work, I am a bit busy at the moment. Can someone show the first couple of lines of work to see if it agrees with mine?

Thanks

 

[topsquark]

Given that $y'=\frac{tan(y)}{1-xsec^2(y)}$, find y'' in terms of $x$ and $y$ only.

I've done this and checked my work several times but my answer does not agree with wolfram alpha. Sorry for not posting my work, I am a bit busy at the moment. Can someone show the first couple of lines of work to see if it agrees with mine?

Thanks

It's straight-forward Calculus, lots of good chain rule exercise, but somewhat appalling in the Algebra.

[math]y'' = \frac{sec^2(y) y' (1 - x~sec^2(y)) - tan(y) (-sec^2(y) - x(2) sec(y) sec(y)~tan(y) y')}{(1 - x~sec^2(y))^2}[/math]

Plug in your value for y' and simplify.

-Dan