What are the derivatives of the inverse trigonometric functions?

#d/dxsin^-1x=1/sqrt(1-x^2)#

#d/dxcos^-1x=-1/sqrt(1-x^2)#

#tan^-1x=1/(1+x^2)#

#cot^-1x=-1/sqrt(1+x^2)#

#sec^-1x=1/(xsqrt(x^2-1))#

#csc^-1x=-1/(xsqrt(x^2-1))#

One useful thing to notice is that the derivatives of all inverse "co" functions are equivalent to the derivatives of the original inverse function, but have a negative added.

Here's a proof for the derivative of the inverse sine function, if you want to avoid memorization. All the other ones can be proved in the same way.

#y=sin^-1x hArr x=siny#, from the definition of an inverse function.

Differentiating #x=siny:#

#d/dx(x)=d/dx(siny)#

#1=cosy*dy/dx# (Implicit Differentiation)

#dy/dx=1/cosy#

We need to get rid of #cosy.# Recall that we said #x=siny# and recall the identity #sin^2theta+cos^2theta=1#. This can be rewritten for #y# and solved for cosine as follows:

#cos^2y=1-sin^2y#

#cosy=sqrt(1-sin^2y)#

Recalling that #x=siny,# then #sin^2y=x^2#

Thus,

#cosy=sqrt(1-x^2)#

#d/dxsin^-1x=1/sqrt(1-x^2)#

I recommend that you commit these integrals to memory- they will help you when you are learning the trig. substitution.