Conversion of one form to another electrostatic integral

I was learning about Electrostatic Potential.

First they introduced,

$$\overrightarrow E(\overrightarrow r) = -\nabla \phi(\overrightarrow r) \quad\quad-(eq \;i)$$

I understood the above equation because both are equal to zero.

But as I was reading further, I came across,

$$-\int^{\overrightarrow r}_{\overrightarrow r_{ref}}\overrightarrow E(\overrightarrow {r'}) \;.\;\overrightarrow {dr'}=\int^{\overrightarrow r}_{\overrightarrow r_{ref} }\nabla\phi(\overrightarrow {r'})\cdot \overrightarrow{ dr'}=\int ^{\overrightarrow r}_{\overrightarrow r_{ref}}\frac{d\phi(\overrightarrow {r'})}{dr'} \;.\;\overrightarrow{dr'}=\phi(\overrightarrow{r'})-\phi(\overrightarrow{r_{ref}})\quad\quad-(eq \;ii)$$

Let me assume eq(ii) this way,

$$I_1=I_2=I_3$$

Now, it is pretty much obvious why $$I_1=I_2$$ (from eq i).

Now, what I am facing difficulty is in figuring out how $$I_2=I_3$$.

If we look closely at $$I_2$$, we find that function of the integral consists of $$\nabla\phi$$, which is nothing but gradient. But as far as I know, $$\nabla$$ consists of partial fraction symbols $$\partial$$. Now, how did $$\partial$$ convert into $$d$$ in $$I_2=I_3$$?

If we closely compare $$I_2$$ and $$I_3$$, we get,

$$\nabla\phi(\overrightarrow{r'})=\frac{d\phi(\overrightarrow{r'})}{dr'}\;.\;\overrightarrow {dr'}$$. How is this part derived? This is my confusion, Can anyone please clarify it (with derivation if possible) ?

\begin{align} \nabla \phi ({\overrightarrow{r'}})\cdot {\overrightarrow{dr'}} &= (\frac{\partial \phi}{\partial x'} \hat{i} + \frac{\partial \phi}{\partial y'} \hat{j} + \frac{\partial \phi}{\partial z'} \hat{k}) \cdot (\vec{dx'} \hat{i} + \vec{dy'} \hat{i} + \vec{dz'} \hat{k}) \\ &= \frac{\partial \phi}{\partial x'} dx'+ \frac{\partial \phi}{\partial y'}dy' + \frac{\partial \phi}{\partial z'} dz' \\ &= d\phi ({\overrightarrow{r'}}) \end{align}
All your integrals are path integrals, meaning that, in more formal terms, you would have to pick a curve $$\gamma$$ joining the two ends and compute
$$\int_0^1\mathbf E(\gamma(t))\cdot\dot\gamma(t)\text dt$$
In general, this integral will depend on the choice of the path $$\gamma$$, but if $$\mathbf E$$ is a conservative field, that is, there exists a scalar field $$\phi$$ such that $$\mathbf E=\nabla\phi$$, then the integral only depends on the ends of the curve $$\gamma(0)$$ and $$\gamma(1)$$.