Question

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

Answer 1

`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!