Find $\frac{d y}{d r}$ for $y=\int_{0}^{r} \sqrt{11+12 t^{2}} d t$

Step 1 ：Given the function \(y=\int_{0}^{r} \sqrt{11+12 t^{2}} d t\), we are asked to find \(\frac{d y}{d r}\).

Step 2 ：This is a problem of differentiation under the integral sign, also known as Leibniz's rule.

Step 3 ：The derivative of the integral of a function from a constant to a variable with respect to that variable is simply the function evaluated at that variable.

Step 4 ：So, we need to find the derivative of the integral of \(\sqrt{11+12t^2}\) from 0 to r with respect to r, which is simply \(\sqrt{11+12r^2}\).

Step 5 ：Using the formula for the derivative of an integral, we find that \(\frac{d y}{d r} = \sqrt{11+12r^2}\).

Step 6 ：However, this can be simplified further.

Step 7 ：By applying the rules of differentiation and simplification, we find that \(\frac{d y}{d r} = \frac{6r^{2}}{\sqrt{12r^{2} + 11}} + \frac{\sqrt{12r^{2} + 11}}{2} + \frac{\sqrt{11}}{2\sqrt{\frac{12r^{2}}{11} + 1}}\).

Step 8 ：Final Answer: \(\boxed{\frac{6r^{2}}{\sqrt{12r^{2} + 11}} + \frac{\sqrt{12r^{2} + 11}}{2} + \frac{\sqrt{11}}{2\sqrt{\frac{12r^{2}}{11} + 1}}}\)