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Consider the following vector function $y: \mathbb R^n \to \mathbb R$

$$ y(\vec x) = y(x_1,x_2,\cdots,x_n)$$

Is it correct to state the following?

$$ dy = \sum_{i=1}^{n}{\left(\frac{\partial y}{\partial x_i}\cdot dx_i\right)} $$

And if so, given the gradient $\nabla y$, defined by

$$ \nabla y = \left( \frac{\partial y}{\partial x_1},\frac{\partial y}{\partial x_2},\cdots,\frac{\partial y}{\partial x_n} \right) $$

Would it also be correct to say this?

$$ dy = \nabla y\cdot d\vec x $$

Much appreciated.

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    $\begingroup$ Absolutely right $\endgroup$ – Shuchang Mar 8 '14 at 3:14
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It is easier to see it from an approximation point of view, let's expand $y(x)$ around some point $x_0$ using Taylor terms, then:

$$y(x)= y(x_0)+\nabla y^T(x-x_0)+(x-x_0)^TH(x-x_0)+hot(x) $$

where $\nabla y$ is the gradient and $H$ is the Hessian matrix (second order derivative) of $y(x)$ both evaluated at $x_0$, and $hot(x)$ are the higher order terms.

Now, if you take the first two terms as an approximation of $y$:

$$y(x)\approx y(x_0)+\nabla y^T(x-x_0)+(x-x_0)^TH(x-x_0)$$

$$\Rightarrow y(x)-y(x_0)\approx \nabla y^T(x-x_0)+(x-x_0)^TH(x-x_0)$$

let $x \rightarrow x_0$, then you have: $$dy \approx \nabla y^Tdx+dx^THdx$$ in the limit $dx^THdx$ goes to zero way faster than $\nabla y^Tdx$, so it's true that:

$$dy =\nabla y^Tdx$$

I'm using a slightly different notation but hope you understand it.

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