Derivative of a quotient involving csc(x)

Apply the quotient rule: if $g(x) = \dfrac{u(x)}{v(x)}$ with $u(x) = \csc(x)$ and $v(x) = x^2$, then $g'(x) = \dfrac{u'(x)v(x) - u(x)v'(x)}{v(x)^2}$. Here $u'(x) = -\csc(x)\cot(x)$ and $v'(x) = 2x$.

Substituting gives $g'(x) = \dfrac{-\csc(x)\cot(x)\cdot x^2 - \csc(x)\cdot 2x}{x^4}$. Factor out $x\csc(x)$ from the numerator to get $\dfrac{x\csc(x)[-x\cot(x) - 2]}{x^4}$, which simplifies to $g'(x) = \dfrac{-\csc(x)\left(x\cot(x) + 2\right)}{x^3}$.

The derivative of cosecant of x is negative cosecant of x times cotangent of x. So the two trig functions that multiply together are cosecant and cotangent. This matches the u prime term used at the start of part a's quotient rule setup.