Question

I have a question on finding the derivative of y. I have attached the question. Thanks for any help!

Answer 1

y = arcsec(1/t)

sec(y) = 1/t

tan^2(y) y' = -1/t^2

s^2 + c^2 = 1
c^2 c^2 c^2
.... t^2 = sec^2-1


(sec^2(y) - 1) y' = -1/t^2

(1/t^2 - 1) y' = -1/t^2

y' = -1/t^2
----------
1/t^2 - 1

multiply by t^2/t^2 and get

y' = -1
---------
1 - t^2

is what im getting

Answer 2

http://www.wolframalpha.com/input/?i=d%2Fdt++arcsec%281%2Ft%29

which means theres an error along the way since sqrt(1-t^2) is correct

Answer 3

derivative of sec = sec tan ....

Answer 4

y = arcsec(1/t)

sec(y) = 1/t

sec(y) tan(y) y' = -1/t^2

1/t tan(y) y' = -1/t^2

tan(y) y' = -1/t

y' = -1/[t tan(y)]

by right triangles ...

Answer 5

hence:

-1/[ t sqrt(1-t^2)/t] = -1/sqrt(1-t^2)

Answer 6

Thank you for the help, amistr64! I see where I went wrong.

Answer 7

i see where i went wrong too :) accursed trig derivatives lol

Answer 8

LOL. amistre64, I see how you went with the problem, but could I not also use the inverse trig function of arcsec which is....look at attached.

Answer 9

of course :) but the general way to prove that derivative is thru the process described. as long as we can remember what the derivative of sec is :)

Answer 10

ok, I was using the inverse trig function when I was stuck with the algebra.

Answer 11

\[\frac{-1/t^2}{\frac{1}{t}\sqrt{\frac{1}{t^2}-1}}\]
\[\frac{-1}{\frac{t^2}{t}\sqrt{\frac{1}{t^2}-1}}\]
\[\frac{-1}{t\sqrt{\frac{1}{t^2}-1}}\]
\[\frac{-1}{\sqrt t^2\sqrt{\frac{1}{t^2}-1}}\]
\[\frac{-1}{\sqrt{t^2(\frac{1}{t^2}-1)}}\]
etc...

Answer 12

Ok, now I see what I did with the way I was doing it. I should have integrated the outside t in the denominator into the square root. Thanks again. You have been great help! I took calc I two years ago and just now getting to Calc II since I'm a full time worker and part time student. Got a little rusty.