By the fundamental theorem of calculus,
[tex]\displaystyle y = \int_0^v \sqrt{3+4t^2} \, dt \implies \boxed{\frac{dy}{dv} = \sqrt{3+4v^2}}[/tex]
In general,
[tex]\displaystyle \frac{d}{dx} \int_0^{g(x)} f(u) \, du = f(g(x)) \frac{dg}{dx}[/tex]
