$Explain partial derivatives v2$

Question

$Explain partial derivatives v2$

I will give you a part of text taken from the book. Please explain the part where author explains directional derivative:
For functions with multiple inputs, we must make use of concept of partial derivatives. The partial derivative $$\frac{\partial}{\partial{x_i}}f(x)$$ measures how f changes as only the variable $$x_i$$ increases at point x. The gradient generalizes the notion of derivative to the case where the derivative is with respect to a vector: the gradient of f is the vector containing all of the partial derivatives, denoted $$\nabla_xf(x)$$
Element i of the gradient is the partial derivative of f with respect to $$x_i$$
Until now everything is clear.
But here comes the part which I don't understand :
The directional derivative in direction u (a unit vector) is the slope of the function f in direction u. In other words, the directional derivative is the derivative of the function $$f(x+\alpha{u})$$with respect to $$\alpha$$
evaluated at $$\alpha = 0$$
Using the chain rule, we can see that $$\frac{\partial}{\partial{\alpha}}f(x+\alpha{u})$$ evaluates to $$u^T\nabla_xf(x)$$ when $$\alpha = 0$$

So please explain:
1. What does $$f(x+\alpha{u})$$ represent (I don't really understand the idea behind x + au)?

2. According to the chain rule we have to have a function where f(x,y) and x is a function of some other variable, together with y, for example x(t), y(t), so f is also a function of t and I can find df/dt. But in case of $$f(x+\alpha{u})$$ I don't really understand why does Chain rule work here, so please explain.

3. And how does derivative of 2 evaluate to $$u^T\nabla_xf(x)$$

P.S. If you spend more time on explaining than equals to 10$ I will tip you.

Calculus

Babaduras

106

Report

Babaduras

0

Is f(x+αu) somehow related to the definition of the of the directional derivative of multivariable function where we take limit of h->0 for (f(x+ha, y+hb) - f(x,y))/h with u = , so f(x+αu) represents f(x+ha, y+hb) where α represents h?
- Babaduras
  
  0
  
  u = (a,b)
- Mathe
  
  +1
  
  Yes, exactly so.

Answer

The answer is accepted.

Join Matchmaticians Affiliate Marketing Program to earn up to a 50% commission on every question that your affiliated users ask or answer.

Babaduras

0

Is f(x+αu) somehow related to the definition of the of the directional derivative of multivariable function where we take limit of h->0 for (f(x+ha, y+hb) - f(x,y))/h with u = , so f(x+αu) represents f(x+ha, y+hb) where α represents h?

Babaduras

0

u = (a,b)

Mathe

+1

Yes, exactly so.

Answer 1

Answers can only be viewed under the following conditions:

The questioner was satisfied with and accepted the answer, or
The answer was evaluated as being 100% correct by the judge.

View the answer

1. $f(x+\alpha u)$ represents the value of the funtion $f$ at the point $x+\alpha u$.

If we wanted to know how much $f$ changes when we move along the direction $u$, starting from the point $x$, we would compute:
$$\lim_{\alpha \rightarrow 0} \dfrac{f(x+\alpha u)-f(x)}{\alpha}$$ Notice that when we compute partial derivatives, we are doing exactly this, only that $u$ in this case would be $(1,\dots,0)$ for the derivative with respect to $x$, or it could be $(0,1,\dots,0)$ for the derivative with respect to $y$, and so on.

2. Notice that $\alpha$ is a scalar that affects all the entries in $x +\alpha u$. For example, in the case of two dimensions, $x +\alpha u = (x_1,x_2)+\alpha(u_1,u_2) = (x_1+\alpha_1 u_1,x_2+\alpha_2 u_2)$

So when we want to understand how $f(x+\alpha u)$ depends on $\alpha$, we must realize that $f$ depends first on its individual entries (so we must compute partial derivatives first), and then compute how does those entries depend on $\alpha$. This is where the expression $$\nabla f (x) \cdot u = u^t \nabla f (x)$$ comes from.

3. The chain rule theorem for functions of several variables states the following:

Let $g: \mathbb{R} \rightarrow \mathbb{R}^n$ and $f: \mathbb{R} ^n \rightarrow \mathbb{R}$ be differentiable functions.

Consider the function $h(\alpha):=f(g(\alpha))$

Then, $h$ is differentiable and we can compute its derivative as $$\dfrac{dh}{d\alpha} = \nabla f(g(\alpha)) \cdot g'(\alpha) $$ This is exactly what is happening above. In our case, $g(\alpha)= x+\alpha u$, for fixed points $x$ and $u$, so that $g'(\alpha) = u$. And this theorem tells us that $f(x+\alpha u)$ is a differentiable function with respect to $\alpha$, and its derivative is given by
$\nabla f(x+\alpha u)\cdot u$.

Mathe

3.6K

Babaduras

+1

I like it, thx.

$Explain partial derivatives v2$

Answer

Related Questions

Search