Question

What is the derivative of `y=csc(x) cot(x)`?

Answer 1

By Product Rule, we can find:
`y'=-csc(x)cot^2(x)-csc^3(x)`

Remember:
`[scs(x)]'=-csc(x)cot(x)`
`[cot(x)]'=-csc^2(x)`

By Product Rule,
`y'=[-csc(x)cot(x)]cdot cot(x)+csc(x) cdot[-csc^2(x)]`,
which simplifies to:
`y'=-csc(x)cot^2(x)-csc^3(x)`