Okay, so...I have a problem with the sub-section of Calculus described above. We have a test tomorrow, but this was only assigned today. I understand, basically, all that we've done before, though.

The instructions are:

Find an equation for the tangent to the graph of y at the indicated point.

1. y = arcsecant(x), x=2

My process:

First, obviously, I found the derivative of the equation and got: y' = 1/(1 + x^2). At two, the slope is 1.

Then, as I remember, I should plug two into the original equation to get the value of "y" at x=2. Theoretically, if I remember correctly, that should give me enough information to finish my problem. However that's where I hit my wall.

I got: y(2) = arcsecant (2) and froze. My calculator gives me an "Error: Domain" message when I try to plug it in and I don't know how to figure it out from the unit circle. Everything I know involves pi in some way or another.

Also, another thing that confused me--though it just may be that the form is unfamiliar to me--was in another problem when the instructions said:

Find f^(-1) and (f^(-1)).

The equation is: f(x) = cosx + 3x

I'm not sure what that means. Do I just write out the reciprocal of that equation for the first part and then find find the derivative of that for the second?

f^(-1) is the inverse of the function. Do you remember how to find the inverse of a function? For an equation like f(x)=cosx+3x, you would solve for x. Do you have to take the derivative of the inverse? You could just take the derivative of the function, then solve for x. I think I repeated myself over and over again. Sorry.

As for the arcsecant problem...

When you find the derivative of the arcsecant, you end up with the equation of the slope. So you would plug the 2 into the derivative to find the slope, not into the original equation.

"If electricity comes from electrons, does morality come from morons?"

For the arcsecant problem, I know that you're supposed to plug two into the derivative to find the slope, but...in order to figure out the equation of the tangent line, aren't you supposed to figure out what y equals as well?

Or...am I overthinking something?

...

Yeah, it's kinda sad. It's been, like, a week and I still need help on the same problem. ><;

First of all, at x=2, i think the slope would be 1/5, not 1. But yeah, you'll need to get the arcsec of 2, but I forgot how...

Actually, I realized just now that I did something incredibly moronic on my part. When I was writing out the equation, I used the derivative formula for arctangent instead of arcsecant. If I had used the correct formula, the slope would have been 1/(2sqrt(3)).

And also, I finally gathered up the courage/time to ask my teacher what I should have done and he explained it to me. It had something to do with unit circles from pre-calculus. I don't really understand it, but I think I've memorized the steps enough..

Anyway, thank you everyone for your help.