Question

need EMERGENCY help for AP Calc BC exam tomorrow.. can't remember how to take derivative of inverse trig functions. (d/dx[sin^-1(x)]) HELP PLEASE!

Answer 1

at this point that's what I'm thinking.

Answer 2

:/ wish i could help

Answer 3

no worries, thanks!

Answer 4

がんばって

Answer 5

thanks, I'm gonna need it.

Answer 6

d/dx[sin^-1(x)] = 1\sqrt[1 - x^2]
Here's the proof:
Let sin^-1(x) = y, therefore x = sin (y).
Differentiate both sides with respect to x and we get
1 = d/dy(sin (y)) dy/dx --> 1 = cos (y) dy/dx --> dy/dx = 1/cos (y)
Recall the elementary trig identity: sin^2 (y) + cos^2(y) = 1.
Solve for cos (y) and we have: cos (y) = sqrt[1 - sin^2(y)]
Recall sin (y) = x therefore cos (y) = sqrt[1 - x^2] and:
d/dx arcsin (x) = 1 / sqrt[1 - x^2]
Hope this helps!! :D

Answer 7

THANK YOU SO MUCH! you're a lifesaver (: