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I dont understand the directional derivative of the function $f(x+ au)$ with respect to a when a =0. According to Goodfellow and al. We can see that, thanks to the chain rule, $\frac{d}{da }f(x+ au)$ evaluates to $u^T\nabla_xf(x)$ when a =0 but

  • when a = 0, isn’t f(x+ au) = f(x) ?
  • if the chain rule is $f(g(x)) = g’(x)f’(g(x))$ why does f(x+ au) gives $\nabla_xf(x)$?

I am a slow learner in mathematics, don’t hesitate to explain it to me as you would with a teenager or with very graphical examples

Revolucion for Monica
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  • It’s really just a case of working it out. Do you know the chain rule for multi-variable functions? When $f(x)$ is a function of $n$ of $n$ variables to the reals, what is the derivative with respect to $x?$ When $g$ is a function of one variable to $n$ variables, what is the derivative of $g?$ – Thomas Andrews Oct 10 '21 at 12:47
  • But, @ThomasAndrews we aren't derivating the f function with respect to x but to a? Shouldn't it be $\nabla_a f(x)$? – Revolucion for Monica Oct 11 '21 at 16:37
  • $\nabla_a f(x)$ doesn’t make any sense, does it, since $f(x)$ doesn’t depend on $a.$ The chain rule for functions of one variable doesn’t do that. If $g(a)=f(x+ua),$ $g’(a)=f’(x+ua)u$ the. $f’$ is the derivative of $f$ here without any usage of the variable $a.$ – Thomas Andrews Oct 11 '21 at 16:47
  • Yes but wern't we starting by deriving with respect to a ($\frac{d}{da}$)? Then how come do we derive/make the gradient with respect to x @ThomasAndrews ? – Revolucion for Monica Oct 12 '21 at 18:11

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