I only just started doing integration and I came across a problem where I was asked to optimise the surface area of a tin can with ribbed sides. $r$ is the radius and $V$ is a constant volume of the tin can. The question is below:
$$A=2\pi r^2+\int_0^{\frac{3V}{2\pi r^2}}\sqrt{4\pi^2r^2+\frac{16\pi^4r^2}{9}\cos^2\left(\frac{20\pi}{3}x\right)}\text{d}x+\frac{V}{2r}$$
Find $$\frac{\text dA}{\text d r}$$
Hint: You can try the Leibniz´s rule to compute the derivative of the term in the middle
$$\frac{d}{dr} \int_{u(r)}^{v(r)}f(r,x) \, dx =f(r,v(r))\cdot \frac{dv}{dr}-f(r,u(r))\cdot \frac{du}{dr}+\int_{_{}^{u(r)}}^{v(r)} \frac{\partial f(r,x)}{\partial r} \, dx$$
with $u(r)=0, v(r)=\frac{3V}{2\pi \cdot r^2}$