# Derivative of $\int_0^tf(x,t)dx$?

Derivative of $$\int_0^tf(x,t)dx$$? Fundamental theorem of calculus works for $$\int_0^tf(x)dx$$

$$\frac{d}{dt}\int_0^tf(x)dx=f(t)$$

• Suppose $F(x,t)$ is an antiderivative of a nice function $f(x,t)$ with respect to $x$. (It is defined up to addition by a function $C(t)$ constant with respect to $x$.) Then $\int_{a(t)}^{b(t)}f(x,t)\mathrm{d}x=F(a(t),t)-F(b(t),t)$, which can be differentiated with the chain rule. Aug 16, 2019 at 2:32
• @runway44 can you give something only about $f$? Aug 16, 2019 at 2:36
• No, in general, you must also differentiate $f$. Try the special case where $f$ doesn't depend on $x$. What's the derivative of $t f(t)$? Aug 16, 2019 at 2:53

$$\frac{d}{dt}\int_0^tf(x,t)dx=\int_0^t\frac{\partial}{\partial t}f(x,t)dx~+~f(t,t)$$
Let $$f(x, t)$$ be a function of $$x$$ and $$t$$ such that both $$f(x, t)$$ and its partial derivative $$\frac{\partial f}{\partial x}$$ are continuous in $$t$$ and $$x$$ in some region of the $$(x, t)$$-plane, including $$a(x) ≤ t ≤ b(x)$$, and $$x_0 ≤ x ≤ x_1$$. Also suppose that the functions $$a(x)$$ and $$b(x)$$ are both continuous and both have continuous derivatives for $$x_0 ≤ x ≤ x_1$$. Then, for $$x_0 ≤ x ≤ x_1$$, $$\frac{d}{dx}\left(\int_{a(x)}^{b(x)} f(x,t) dt\right)=\int_{a(x)}^{b(x)} \frac{\partial }{\partial x}f(x,t) dt +f( x, b(x)) \frac{db}{dx}-f( x, a(x)) \frac{da}{dx}$$
• Can I have something only in terms of $f$? But not derivative of $f$? Aug 16, 2019 at 2:33
• No, because for this case $~f~$ is a function of two variable $~x,~t~$. If you take $~f~$ is function of one variable, then Leibniz Integral Rule becomes Fundamental theorem of calculus. Aug 16, 2019 at 2:39