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

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Answer 1 · 2015-05-21T16:47:29

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

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

The chain rule states that

#(dy)/(dx)=(dy)/(du)(du)/(dx)#

#(dy)/(du)=-u^-2#

#(du)/(dx)=-1*cscxcotx-1*csc^2x=-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#