How do you differentiate #f(x)=sqrt(1/csc(2x -4) # using the chain rule?

1 Answer
Nov 29, 2016

Answer:

#dy/dx= (cot(2x- 4)csc(2x- 4))/(csc(2x - 4))^(3/2)#

Explanation:

First of all, this can be written as #f(x) = 1/sqrt(csc(2x- 4))#, which in turn can be written as #f(x) = (csc(2x- 4))^(-1/2)#.

To differentiate this function, we will apply the chain rule--twice.

We let #y = u^(-1/2), u = cscv# and #v = (2x - 4)#.

By the chain rule, #dy/dx= dy/(du) xx (du)/(dv) xx (dv)/dx#, therefore we will have to differentiate each function.
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#dy/(du) = -1/2u^(-3/2) = -1/(2u^(3/2))#

#(du)/(dv) = (0 xx sinv - cosv xx 1)/(sinv)^2 = -cosv/sin^2v= -cotvcscv#

#(dv)/dx = 2#
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#dy/dx = dy/(du) xx (du)/(dv) xx (dv)/dx#

#dy/dx= -1/(2u^(3/2)) xx -cotvcscv xx 2#

#dy/dx= (2cot(2x- 4)csc(2x - 4))/(2(csc(2x- 4))^(3/2))#

#dy/dx= (cot(2x- 4)csc(2x- 4))/(csc(2x - 4))^(3/2)#

Hopefully this helps!