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.

  • 1
    $\begingroup$ Absolutely right $\endgroup$ – Shuchang Mar 8 '14 at 3:14

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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.