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October 18, 2007 - 22:13
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Jayne Do [3]
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Problem with Derivatives of Inverse Trigonometric Functions

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?

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[1] https://course-notes.org/forum/math/ap_calculus/problem_with_derivatives_of_inverse_trigonometric_functions#comment-1062447 [2] https://course-notes.org/forum/math/ap_calculus/problem_with_derivatives_of_inverse_trigonometric_functions [3] https://course-notes.org/user/59733