How do you find the derivative of #y=arcsin(1/x)#?
2 Answers
First, recall the identity
If this identity doesn't look familiar then I may recommend viewing a few videos from this page as they present a couple identities like this, and explain why they are true.
Differentiating
The derivative of
Now, all we need to do is simplify a bit:
Explanation:
We may also know from the outset that:
#d/dx"arcsec"(x)=1/(absxsqrt(x^2-1))# #d/dx"arccsc"(x)=-1/(absxsqrt(x^2-1))#
Then:
#y=arcsin(1/x)#
#sin(y)=1/x#
#1/(sin(y))=x#
#csc(y)=x#
#y="arccsc"(x)#
Then:
#dy/dx=-1/(absxsqrt(x^2-1))#