What is the derivative of (cscx+cotx)^-1(cscx+cotx)1?

1 Answer
May 21, 2015

Let's use chain rule here, by naming u=(cscx+cotx)u=(cscx+cotx)

Thus, our function is written as y=u^-1y=u1.

The chain rule states that

(dy)/(dx)=(dy)/(du)(du)/(dx)dydx=dydududx

(dy)/(du)=-u^-2dydu=u2

(du)/(dx)=-1*cscxcotx-1*csc^2x=-csc(cotx+cscx)dudx=1cscxcotx1csc2x=csc(cotx+cscx)

Joining them both:

(dy)(dx)=(cancel(-)u^-2)/(cancel(-)csc(cotx+cscx))

Now, substituting u:

(dy)/(dx)=cancel(cscx+cotx)/(cscxcancel(cscx+cotx))=1/cscx