Derivative of an integral function Say $$\varphi(\theta) = \int_{0}^{2\pi} \frac{f((1-\theta)z+\theta(z_0+re^{it}))}{z_0+re^{it}-z}ire^{it}dt.$$ Why do we have $$\varphi'(\theta) = \int_{0}^{2\pi} f'((1-\theta)z+\theta(z_0+re^{it}))ire^{it}dt?$$ What theorem allows me to say that the derivative of the integral is the derivative of the integrant?

Note that

$$\frac{d}{dx} \left (\int_{a(x)}^{b(x)}f(x,t)dt \right) = f(x,b(x))\cdot b'(x) - f(x,a(x))\cdot a'(x) + \int_{a(x)}^{b(x)} \frac{\partial}{\partial x}f(x,t) dt$$

also known as _differentiation under the integral_.

In your case $a$ and $b$ are both costants, so the first part of the RHS is just zero and you'll get:

$$\frac{d}{dx} \left (\int_{a}^{b}f(x,t)dt \right) = \int_{a}^{b} \frac{\partial}{\partial x}f(x,t) dt$$

This is known as the _Leibnitz rule_.