How do you differentiate #f(x)=sqrtcsc(2x -4) # using the chain rule?

Question

1415 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-02-09T01:51:15

#f'(x)=-cos(2x-4)csc(2x-4)^(3/2)#

rewrite the function like this
#csc(2x-4)^(1/2)#

use chain rule by first using power rule on the #1/2# exponent,
then on the csc function, and finally on the function inside the csc

#cancel(1/2)csc(2x-4)^(-1/2)*(-csc(2x-4)cot(2x-4))cancel(2)#

then

#csc(2x-4)^(-1/2)=1/sqrt(csc(2x-4))#

multiply

#(-csc(2x-4)cot(2x-4))/(sqrt(csc(2x-4))#

convert everything into sin and cos
#((-1/sin(2x-4))*(cos(2x-4)/sin(2x-4)))/csc(2x-4)^(1/2)#

simplify

#((-cos(2x-4))/(sin^2(2x-4)))/(1/sin(2x-4)^(1/2))#

flip and multiply

#(-cos(2x-4)sin(2x-4)^(1/2))/sin^2(2x-4)#

subtract sin exponents

#-cos(2x-4)*sin(2x-4)^(-3/2)#

more simplification

#-cos(2x-4)*(1/csc(2x-4)^(-3/2))#

#-cos(2x-4)csc(2x-4)^(3/2)#