Is there an easy way to memorize the derivatives of inverse trigonometric functions?

2 Answers
Apr 7, 2015

The way is not to memorize.
The easiest way is to derive the formulae.
For e.g

#y=cos^-1(x)#
then
#x=cosy#
#dx = -siny dy#
# dy/dx = -1/sin(y)#
#dy/dx =-1/sqrt(1-x^2)#

Apr 10, 2015

There is some way to memorize, though you shouldn't have to. Knowing how to derive it is helpful for figuring them out again (maybe your internet goes out... I dunno). You can do it in "regular" and "co-" pairs.

While #(d(arc tanx))/dx = 1/(1+x^2)#,

#(d(arc cotx))/dx = -1/(1+x^2)#.


While #(d(arc sinx))/dx = 1/sqrt(1-x^2)#,

#(d(arc cosx))/dx = -1/sqrt(1-x^2)#.


While #(d(arc secx))/dx = 1/(absxsqrt(x^2-1))#,

#(d(arc cscx))/dx = -1/(absxsqrt(x^2-1))#.

Now you only have to remember three of them. "co-" implies a reversed sign.