Using the definition of derivative to find $\tan^2x$ The instructions: Use the definition of derivative to find $f'(x)$ if $f(x)=\tan^2(x)$. 
I've been working on this problem, trying every way I can think of. At first I tried this method:
$$\lim_{h\to 0} {\tan^2(x+h)-\tan^2(x)\over h}$$
$$\lim_{h\to 0} {\tan(x+h)-\tan(x)\over h}\cdot\lim_{h\to 0} {\tan(x+h)-\tan(x)\over h}$$
And then I went on from there, but I was never able to get rid of the $h$.
So then I tried this:
$$\lim_{x\to y} {\tan(x)-\tan(y)\over x-y}\cdot\lim_{x\to y} {\tan(x)-\tan(y)\over x-y}$$
$$\lim_{x\to y} {\tan(x-y)[1+\tan(x)\tan(y)]\over x-y}\cdot\lim_{x\to y} {\tan(x-y)[1+\tan(x)\tan(y)]\over x-y}$$
$$\lim_{x\to y} {\sin(x-y)\over (x-y)}\cdot{1\over \cos(x-y)}\cdot[1+\tan(x)\tan(y)]\cdot\lim_{x\to y} {\sin(x-y)\over (x-y)}\cdot{1\over \cos(x-y)}\cdot[1+\tan(x)\tan(y)]$$
I then put the ${\frac{\sin(x-y)}{x-y}}=1$, and I traded all of the $y$ values for $x$, which gave me $\frac{1}{\cos(\theta)}=1$
$$1+\tan^2(x)\cdot1+\tan^2(x)=1+2\tan^2(x)+\tan^4(x)$$
I know that the derivative of $\tan^2(x)=2\tan(x)\sec^2(x)$, so this is obviously wrong. 
I found this link from a previous stack exchange point on finding the limit by definition of $\tan(x)$, so I tried using the answers given there. But, even with that, I haven't been able to get this correct.
What do I need to do?

i think your derivative (with brackets) is ok since $\tan(x)$ has a derivative thus $\tan(x)\tan(x)=\tan(x)^2$ has also a derivative and is given by $2\cdot(1+\tan(x)^2)(\tan(x))$ (in the range of definition) a comment before the downvoting is starting: $$\tan(x)'=\frac{1}{\cos(x)^2}=\frac{\sin(x)^2+\cos(x)^2}{\cos(x)^2}=1+\tan(x)^2$$
