Answer: (i) \(4\tan x+4x\operatorname{sec}^2x\). (ii) \(\dfrac{(x^2-1)(3e^{3x+1})-2xe^{3x+1}}{(x^2-1)^2}\), equivalently \(\dfrac{e^{3x+1}(3x^2-2x-3)}{(x^2-1)^2}\).

(i) Use the product rule on \(4x\tan x\):

\(\frac{d}{dx}(4x\tan x) =4\tan x+4x\operatorname{sec}^2x.\)

(ii) Use the quotient rule. Let \(u=e^{3x+1}\) and \(v=x^2-1\). Then

\(u'=3e^{3x+1}, \qquad v'=2x.\)

Therefore

\(\frac{d}{dx}\left(\frac{e^{3x+1}}{x^2-1}\right) = \frac{(x^2-1)(3e^{3x+1})-2xe^{3x+1}}{(x^2-1)^2}.\)

This can also be written as

\(\frac{e^{3x+1}(3x^2-2x-3)}{(x^2-1)^2}.\)