Question

Differentiate
ln cosec(x- pi)

Answer 1

looks like 3 layers to undo

Answer 2

ln' cosec(x- pi) * cosec'(x- pi) * (x- pi)'

Answer 3

think of it as:

ln'a * csc' b * x'

Answer 4

So, how would you solve it?

Answer 5

dy/dx = [1/cosec(x-pi) ]* [ - cosec(x-pi)* cot (x-pi) ] as differentiation of cosecx is -cosecxcotx.

Answer 6

The answer = -cotx?

Answer 7

Yup.

Answer 8

that IS how you would solve it :)

Answer 9

How?

Answer 10

answer will be -cot(x-pi)

Answer 11

- think the -pi is spurious

Answer 12

The book just says -cotx. How do you get it? I still don't understand...?

Answer 13

ln csc(x- pi) = ln -csc(x)

Answer 14

yea because -cot( x- pi) can be written as cot(pi-x) which is equal to -cotx.

Answer 15

which just goes thru the loops

-csc(x)'
------ * -csc(x)'
-csc(x)

Answer 16

i might of an extra part in that :)

Answer 17

Thanks :) I understand now.

Answer 18

:) yw

Answer 19

>> http://www.wolframalpha.com/bing/?i=derivative+of+ln+csc(x-+pi)&expl=3

Answer 20

There's one other thing worth observing here.

As \[ \ln(1/x) = -\ln x, \] it follows that
\[ \ln \csc(x - \pi) = -\ln \sin(x-\pi) \]
This now makes the differentiation a little more straight-forward.