= Total derivative
{title2=$\frac{dF}{dt}=\frac{\partial F}{\partial t}+\sum_i\frac{\partial F}{\partial x_i}\frac{dx_i}{dt}$}
{wiki}
A <total derivative> accounts for every dependence of a quantity on the variable of <differentiation>. For a <differentiable function> $F(t,x_1,\ldots,x_n)$ evaluated along <differentiable> coordinate <functions> $x_i(t)$, the <multivariable chain rule> gives
$$
\frac{d}{dt}F(t,x_1(t),\ldots,x_n(t))
=\frac{\partial F}{\partial t}+\sum_{i=1}^n\frac{\partial F}{\partial x_i}\frac{dx_i}{dt}.
$$
All the <partial derivatives> on the right are evaluated at $(t,x_1(t),\ldots,x_n(t))$. The first term captures explicit dependence on $t$; the remaining terms capture changes transmitted through the coordinates.
For example, take $F(t,x)=tx^2$ and $x(t)=t$. The <partial derivative> with respect to $t$, holding $x$ fixed, is $x^2$, whose value on the path is $t^2$. The <total derivative> is instead
$$
\frac d{dt}F(t,t)=\frac d{dt}t^3=3t^2
=\underbrace{t^2}_{\partial_t F}+\underbrace{2t^2}_{\partial_x F\,x'}.
$$
At $t=2$, the <partial derivative> is $4$ and the <total derivative> is $12$. The difference is the contribution from changing $x$.
If the composite <function> is <continuously differentiable> on $[a,b]$, the <fundamental theorem of calculus> gives
$$
\int_a^b\frac{d}{dt}F(t,x(t))\,dt=F(b,x(b))-F(a,x(a)).
$$
This is the sense in which an integrated <total derivative> contributes endpoint terms. The endpoint statement concerns the whole composite <function> along the path. https://openstax.org/books/calculus-volume-3/pages/4-5-the-chain-rule[OpenStax, Calculus Volume 3: the chain rule with dependent variables]{ref}.
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