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