About the formula for the magnitude of the gradient of a scalar valued function, $|\nabla T|$ Let $T: R^3 \rightarrow R$. I recognize that $|\nabla T|$ is the rate of change of $T$ in the direction of the most rapid increase.
Say we move by $d\vec{l} = (dx, dy, dz)$. When $d\vec{l}$ is parallel to $\nabla T$, then the dot product is: $\nabla T \cdot d\vec{l} = |\nabla T|\times |d\vec{l}|$, because in this case $\cos \theta = 1$.
We know that $\nabla T \cdot d\vec{l} = dT$, so combining these facts we get: $dT = |\nabla T|\times |d\vec{l}|$.
Then we get $$|\nabla T| = \frac{dT}{|d\vec{l}|}$$ which is quite obvious.
We know that $|d\vec{l}| = \sqrt{dx^2 +dy^2 +dz^2}$, so the above expression becomes:
$$|\nabla T| = \frac{dT}{|d\vec{l}|} = \frac{dT}{\sqrt{dx^2 +dy^2 +dz^2}}$$
But from the definition of the gradient, $$\nabla T = \left(\frac{\partial T}{\partial x},\frac{\partial T}{\partial y}, \frac{\partial T}{\partial z}\right)$$
we know that $$|\nabla T| = \sqrt{\left(\frac{\partial T}{\partial x}\right)^2+\left(\frac{\partial T}{\partial y}\right)^2+\left(\frac{\partial T}{\partial z}\right)^2}$$
But this does not match with the intuitive formula above, $$|\nabla T| = \frac{dT}{\sqrt{dx^2 +dy^2 +dz^2}} = \sqrt{\frac{dT^2}{dx^2 +dy^2 +dz^2}}$$
Why is this intuition wrong?
 A: They actually do match!
Recall that we have $\mathrm{d}T=\dfrac{\partial T}{\partial x}\mathrm{d}x+\dfrac{\partial T}{\partial y}\mathrm{d}y+\dfrac{\partial T}{\partial z}\mathrm{d}z$ in general. Also, note that if $\mathrm{d}\vec{\ell}=\left\langle \mathrm{d}x,\mathrm{d}y,\mathrm{d}z\right\rangle$  is the direction of greatest increase, then it's parallel to $\nabla T=\left\langle \dfrac{\partial T}{\partial x},\dfrac{\partial T}{\partial y},\dfrac{\partial T}{\partial z}\right\rangle$ . Set $r$ to be positive so that $r\,\mathrm{d}\vec{\ell}=\nabla T$.
Then we have \begin{align*}
&\phantom{=}\dfrac{\mathrm{d}T}{\sqrt{{\mathrm{d}x}^{2}+{\mathrm{d}y}^{2}+{\mathrm{d}z}^{2}}}\\&=\dfrac{\dfrac{\partial T}{\partial x}\mathrm{d}x+\dfrac{\partial T}{\partial y}\mathrm{d}y+\dfrac{\partial T}{\partial z}\mathrm{d}z}{\sqrt{{\mathrm{d}x}^{2}+{\mathrm{d}y}^{2}+{\mathrm{d}z}^{2}}}\\&=\dfrac{r{\mathrm{d}x}^{2}+r{\mathrm{d}y}^{2}+r{\mathrm{d}z}^{2}}{\sqrt{{\mathrm{d}x}^{2}+{\mathrm{d}y}^{2}+{\mathrm{d}z}^{2}}}\\&=r\sqrt{{\mathrm{d}x}^{2}+{\mathrm{d}y}^{2}+{\mathrm{d}z}^{2}}\\&=\sqrt{\left(r\mathrm{d}x\right)^{2}+\left(r\mathrm{d}y\right)^{2}+\left(r\mathrm{d}z\right)^{2}}\\&=\sqrt{\left(\dfrac{\partial T}{\partial x}\right)^{2}+\left(\dfrac{\partial T}{\partial y}\right)^{2}+\left(\dfrac{\partial T}{\partial z}\right)^{2}}\end{align*}
Mathematical Meaning
If you're concerned about the manipulations above, there are lots of ways of making them (or slight variations of them) valid/rigorous. The simplest would be to say that everything above is approximately equal when $T$ is differentiable and things like $\mathrm dT$ and $\mathrm dx$ are tiny real numbers.
