Question

how do you take the derivative of inverse trig?

Answer 1

use implicit method

Answer 2

\[y=trig^{-1}(x)\]
=>
\[trig(y)=x\]
\[y' (trig')(y)=1 => y'=\frac{1}{(trig')(y)}\]

Answer 3

for example y=arcsin x
then x=siny
d/dx x = d/dx siny
notice that dsiny=dy cos y
so dy/dx=cosy=cos arcsin x =1/sqrt(1-x^2)

Answer 4

(trig')(y) is not multiplication by the way

Answer 5

sorry i made a mistake,
it should be dy/dx=1/cosy=1/cos arcsin x

Answer 6

here's a link:
http://tutorial.math.lamar.edu/Classes/CalcI/DiffInvTrigFcns.aspx