Question

find the derivative y=sec^2x - tan^2x

Answer 1

Oh this is a silly problem :)
If you remember your trig identities,
sec^2x=1+tan^2x.. so we really have y=1 after you cancel out the tangents.

So we know the answer SHOULD BE 0 after taking a derivative, but I guess we'll need to show the steps. heh :D

What are you having trouble with? are the squares confusing you? :o

Answer 2

\[\large y=\sec^2x-\tan^2x\]\[\huge y'=2\sec^1x(\sec x)'-2\tan^1 x(\tan x)'\]Understand how I did that? I applied to the power rule to the OUTER function (the square), then by the chain rule, we have to multiply by the derivative of the inside.

The little prime is to show that we still need to differentiate that part.

Answer 3

so the answer is 0?

Answer 4

Eventually, yes -_- but I think they want you to take the steps and actually take the derivative :P

Answer 5

how do i do this step by step?

Answer 6

The first step is already shown, taking the derivative of the outermost function, then multiplying by the derivative of the inside.

If we take the derivative of the inner function (the ones with primes on them), we get,\[\large y'=2\sec x(\sec x \tan x)-2\tan x(\sec^2x)\]You need to remember a few derivatives to get through this problem. Namely, the derivative of secant and that of tangent.

Answer 7

the derivative of secant is secx*tanx and the derivative of tangent is sec^2(x)

Answer 8

Yes good c: