Finding derivative of given function. f(t) = $\int_{t^2}^{4}\sqrt{\cos(x)+12}\;dx$
Rearranging limits of integration...
$f(t) = -\int_{4}^{t^2}\sqrt{\cos(x)+12}\;dx$
Taking derivative...
$f'(t) = -\sqrt{\cos(t^2)+12}\; - (1/2)({\cos(t^2)+12})^{-1/2}\; - \sin{(t^2)} \cdot 2t$
Is this answer correct, or am I taking "too much" of the derivative? Should I only be taking the derivative of $t^2$? Could someone please explain this?

Yes, you "took too much derivative" here:
We need only apply the chain rule: If
$$F'(t)= - \int_4^{\large u(t) = t^2}\sqrt{\cos x + 12} \, dx $$ $$\implies F(t) = -\Big[\underbrace{t^2}_{\large u(t)}\Big]' \sqrt{\cos (t^2) + 12} = -\underbrace{2t}_{\large u'(t)}\sqrt{\cos (t^2) + 12}$$
