Derivative of an accumulation function with a variable upper limit

By FTC Part 1, if $F(x) = \int_a^{g(x)} f(t)\,dt$, then $F'(x) = f(g(x)) \cdot g'(x)$. Here $f(t) = \dfrac{1}{1+t^2}$ and the upper limit is $g(x) = x^3$, so we need both $f(g(x))$ and $g'(x)$.

Substituting $t = x^3$ into $f(t)$ gives $f(g(x)) = \dfrac{1}{1+(x^3)^2} = \dfrac{1}{1+x^6}$. Since $g'(x) = 3x^2$, multiplying gives $F'(x) = \dfrac{3x^2}{1+x^6}$.