Simplify $\frac{d}{dt}\int_x^t f(t,y)dy$

Question

I am trying to simplify $\frac{d}{dt}\int_x^t f(t,y)dy$ as a part of a proof.

I am somewhat confused on how I can proceed with this. Do I define a function $g(t,y)$ such that $\frac{\partial g}{\partial y} = f(x,t)$ and then say $\frac{d}{dt}\int_x^t f(t,y)dy = \frac{d}{dt}(g(t,t) - g(t,x))$?

Or can I just say that $\frac{d}{dt}\int_x^t f(t,y)dy = \frac{d}{dt}(f(t,t) - f(t,x))$? If so why can I say this.

Also note that I am trying to avoid using leibniz integral rule since it has not been covered yet.

Answer 1

You can think of the integral as being a function of three parameters:

$$g(a,b,c) = \int_a^b f(c,y)\:dy$$

Thus the derivative you want can be derived from chain rule

$$\frac{d}{dt}g(a(t),b(t),c(t)) = \frac{\partial g}{\partial a}\frac{da}{dt} + \frac{\partial g}{\partial b}\frac{db}{dt} + \frac{\partial g}{\partial c}\frac{dc}{dt}$$

$$= 0 + f(t,t) +\int_x^t \frac{\partial f}{\partial t}(t,y)\:dy $$