# How to prove $\nabla\cdot \vec{B}=0 \Rightarrow \exists \vec{A}:\vec{B}=\nabla \times \vec{A}$

Suppose $\vec{B}$ is a differentiable vector field defined everywhere such that $\nabla\cdot \vec{B}=0$. Define $\vec{A}$ by the integral $$A_1=\int_0^1 \lambda(zB_2(\lambda x,\lambda y,\lambda z)- yB_3(\lambda x,\lambda y,\lambda z)) d\lambda$$ Together with its two cyclic permutations for $A_2,A_3$

I'm trying to work out two things here:

$1.$ What is $\frac{d}{d\lambda}B_i(\lambda x,\lambda y,\lambda z)$

$2.$ How we can use $1.$ to determine $\frac{\partial A_2}{\partial x}-\frac{\partial A_1}{\partial y}=B_3$

From this we can deduce the existance of the magnetic potential by extending $2.$, this is what I have so far:

Is $\frac{d}{d\lambda}B_i(\lambda x,\lambda y,\lambda z)=(x,y,x) \cdot \nabla B_i$? And can we bring the partial derivative on $A_i$ inside the integral? I have proceeded along these lines but have not found a way to substitute.

Any help would be greatly appreciated!

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