Question

find the derivative of h(x)=2x^4cotx/3x^2
is there a better way to do this using the quotient rule? this feels wrong

Answer 1

If you want to avoid using the quotient rule, you can cancel the powers of x :

$h(x) = \frac{2x ^{4} \cot(x) }{3 {x}^{2} } = \frac{2}{3} {x}^{2} \cot(x)$

Then differentiate using the product rule.

$h'(x)=\frac23\left(2x\cot(x)-x^2\csc^2(x)\right)$

$h'(x)=\frac43x\cot(x)-\frac23x^2\csc^2(x)$