= Solution
For $x>0$, the kernel is smooth between the two positive endpoints. The <Leibniz integral rule> gives both endpoint terms and the parameter derivative:
$$
I'(x)=\frac{3\cos(x^4)-2\cos(x^3)}x-\int_{x^2}^{x^3}\sin(xt)dt.
$$
The remaining integral equals $[\cos(x^3)-\cos(x^4)]/x$, so
$$
\boxed{I'(x)=\frac{4\cos(x^4)-3\cos(x^3)}x,\qquad x>0.}
$$
The substitution $u=xt$ gives the same result directly from limits $x^3,x^4$. Negative $x$ would make the original interval cross the nonintegrable $1/t$ singularity, so it is not part of this ordinary-integral domain.
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