f(x)=sec

^{2}x and g(x)=tan

^{2}x. Show that f'(x)=g'(x)

Well I didn't know how to do secant or tangent very well.

So I converted it into sine and cosine.

sec

^{2}x = 1/cos

^{2}x

Then I took the derivate of cosine

f'(x)=1/-sin

^{2}x

Then somehow I think I'm supposed to get g(x) into that as well?

Can someone explain it to me so I could understand it better. **P.S.** Also does anyone know of a good website for trigonometri functions that convert sine, cosines, tagent, etc.

Like for example, like sinx/x = 1, or 1-cosx/cox=0, then there are a bunch of other rules. Anyone know the website of that? I used to have them in my notes, but then I lost the paper