Question

d/dt sec^(-1) (1/2t^3)

Answer 1

Consider two skills you already have. The ability to find the derivative of normal trig functions and implicit differentiation. So you can turn this around from:

y=sec^-1(1/2t^3) into sec(y)=1/2t^3 and you might find this easier!

Answer 2

how do you go from sec^-1 to sec^1 ??

Answer 3

Sec^-1 is the inverse secant function, also called arcsec. It's the reverse of secant. So what you can do is take the sec( to both sides to get rid of the arcsec. Similar to how you can square both sides of a function to get out of a square root.

Answer 4

sec (sec^1 1/(2t^3))

is that what you mean?

Answer 5

sec(sec^-1(x))=x

Answer 6

use chain rule
let u = 1/2t^3
y = sec^-1 u

dy/du = 1 / ( sqrt(1 + (u/2)^2) u)
and u = (1/2)* t^-3
du = (-3/2) * t^(-4)

dy/dt = dy/du * du/dt

Answer 7

cwrw this won't work because this isn't secant to the -1 power this is arcsecant.

"The notations sin−1, cos−1, etc. are often used for arcsin, arccos, etc., but this convention logically conflicts with the common semantics for expressions like sin2(x), which refer to numeric power rather than function composition, and therefore may result in confusion between multiplicative inverse and compositional inverse." - Wikipedia

Answer 8

sec(sec^-1(x)) doesnt work..

Answer 9

http://www.wolframalpha.com/input/?i=sec(sec%5E-1(x))&t=crmtb01

Answer 10

Yeah but how can I evaluate d/dt using that expression?

Answer 11

Have you learned how to take the derivative of trigonometric functions and have you learned to do implicit differentiation yet?

Answer 12

Yes