Im not sure if there is a typo, or I'm just not seeing how to solve this:
help please (:
Compute the derivative w.r.t. t of g(t) := ∫_{0}^{t } f(t,x) dx.
sorry for the iffy notation
Im not sure if there is a typo, or I'm just not seeing how to solve this:
help please (:
Compute the derivative w.r.t. t of g(t) := ∫_{0}^{t } f(t,x) dx.
sorry for the iffy notation
Use the definition of the derivative:
$\displaystyle \frac{\partial}{\partial\,t}\int_0^t f(t,\,x)\, \operatorname{d}x \,=\, \lim_{h\to 0} \frac{\int_0^{t+h} f(t+h,\,x)\, \operatorname{d}x - \int_0^t f(t,\,x)\, \operatorname{d}x}{h}$
You see that the first term in the fraction can be written as
$\displaystyle \int_0^{t+h} f(t+h,\,x)\, \operatorname{d}x \,=\, \int_0^t f(t+h,\,x)\, \operatorname{d}x + \int_t^{t+h} f(t+h,\,x)\, \operatorname{d}x$
$\displaystyle =\, \int_0^t f(t+h,\,x)\, \operatorname{d}x + h\,f(t,\,t) + O(h^2)$
$\displaystyle =\, \int_0^t \left(f(t,\,x) + h\,\frac{\partial}{\partial t}f(t,\,x) + O(h^2)\right) \operatorname{d}x + h\,f(t,\,t) + O(h^2)$
From there it should be an easy task to escape the limit formulation and obtain the derivative.
Good luck!
I would recommend "Leibniz' rule" which generalizes the "Fundamental Theorem of Calculus":
$\displaystyle \frac{d}{dx}\int_{u(x)}^{v(x)} f(x,t)dt= f(x, u(x))- f(x,v(x))+ \int_{u(x)}^{v(x)}\frac{\partial f}{\partial x}dx$
I think that should be
$\displaystyle \frac{d}{dx}\int_{u(x)}^{v(x)} f(x,t)dt = \frac{dv(x)}{dx} f(x,v(x)) - \frac{du(x)}{dx} f(x, u(x)) + \int_{u(x)}^{v(x)}\frac{\partial f(x,t)}{\partial x}dt$