Evaluating the commutator of derivative and position

Question

In Zettili's book on quantum, the fully worked problem 2.6 asks to show $$ \hat{A} = i(\hat{X}^2+1)\frac{d}{dx} + i\hat{X}. $$ Is Hermitian. Where $\hat{X}$ is the position operator. I took the hermitian conjugate of $\hat{A}$ and end up with the result $$ \hat{A}^{\dagger} = i\hat{X}^2\frac{d}{dx}+i\left[\frac{d}{dx},\hat{X}^2\right]+i\frac{d}{dx}-i\hat{x} $$ but when I go to compute the commutator, I have to calculate $$ \left[\frac{d}{dx},\hat{X}\right] $$ using $[A,B] = AB-BA$ I get $$ \left[\frac{d}{dx},\hat{X}\right]=\frac{d}{dx}(x)-x\frac{d}{dx} =1-x\frac{d}{dx}. $$ Though in the solution provided in the textbook, they use $\left[\frac{d}{dx},\hat{X}\right]=1$. I'm hoping I'm making a stupid error somewhere but I can't seem to find it. Any nudge in the right direction would be greatly appreciated.