Answer
$u^{\prime}(x)= (6x+1) \cos (3x^2+x-1)$
Work Step by Step
We have: $u(x)= \sin (3x^2+x-1)$
We differentiate both sides with respect to $x$.
$u^{\prime}(x)=\dfrac{d}{dx} [ \sin (3x^2+x-1)] \\= \cos (3x^2+x-1) \times (6x+1)$
Simplify to obtain:
$u^{\prime}(x)= (6x+1) \cos (3x^2+x-1)$
