How do you use the chain rule to differentiate #y=csc^3((3x)/2)#?

1 Answer
Jan 30, 2018

#d/dx( csc^3((3x)/2)) = -9/2 * csc^3((3x)/2) * cot((3x)/2) #

Explanation:

#d/dx f(g(x))^n = n * f(g(x))^(n-1) * d/dx f(g(x)) * d/dx g(x)#

#d/dx csc(x) = -csc(x) cot(x)#

#d/dx (3x)/2 = 3/2#

Applying the first formula:

#d/dx( csc^3((3x)/2)) = 3* csc^2((3x)/2)* (-csc((3x)/2)*cot((3x)/2))*3/2#

Simplifying:

  • #3*3/2 = 9/2#
  • #csc^2((3x)/2) * (-csc((3x)/2)) = -csc^3((3x)/2)#

We end with:

#d/dx( csc^3((3x)/2)) = -9/2 * csc^3((3x)/2) * cot((3x)/2) #

Hope that helped!