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!