If f(x,y,z)= $\int_{c} (F.dl)$ ,to prove that grad f= F is used the path from the origin to a generic point (x,y,z). And first we move along the x axis then y axis then z axis.therefore f(x,y,z)= $\int_{c} (F.dl)=\cdots=\int_{0}^{x} (F_1(t,0,0)dt)+\int_{0}^{y} (F_2(x,t,0)dt)+\int_{0}^{z} (F_3(x,y,t)dt)$
Where $\F=(F_1,F_2,F_3)\$
Now using the fundamental theorem of calculus it can be show that $\frac{\partial f}{\partial x}=F_1(x,y,z)$ the same for y and z. After this is easy to prove grad f= F
My question is how we use the fundamental theorem of calculus to prove that $\frac{\partial f}{\partial x}=F_1(x,y,z)$?for the others cases will be the same,can someone help me for the first derivative of x?
I try some ways but arrive nowhere,can anyone please help me on this one?
I can see that intuitively but I can`t express that mathematicly
$$\int_{c} F\cdot ds=\int_{0}^{x} F_1(t,0,0)dt+\int_{0}^{y} F_2(x,t,0)dt+\int_{0}^{z} F_3(x,y,t)dt $$
Using this representation of $F$, you can see that $\dfrac{\partial F}{\partial z}=F_3$, since the first two terms are independent of $z$, and the last term is of the form $\displaystyle \int_0^z G(t)dt$ where $G(z)=F(x,y,z)$. Since $x,y$ are just constants, you're done, since by FTC $\displaystyle \frac{d}{dz}\int_0^z G(t) dt=G(z)$