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Please can somebody check my answer? Tell me and explain me my mistakes and so on if there is. Thank you for helping :)

Question:

Suppose that the function $f:\Bbb R^n \to \Bbb R$ is continuously differentiable.

Let $x$ be a point in $\Bbb R^n$. For $p$ a nonzero point in $\Bbb R^n$ and $\alpha$ be a nonzero real number. Show that

$\dfrac{\partial f}{\partial(\alpha p)}(x) = \alpha \dfrac{\partial f}{\partial p}(x)$

Solution:

$\dfrac{\partial f}{\partial(\alpha p)}(x)=$

$\displaystyle = \lim_{t \to 0}\left(\frac {f(x+\alpha tp)-f(x)}{t}\right)$ by the definition of directional derivative of $f$

$\displaystyle =\lim_{t\to 0}\left(\sum_{i=1}^{n} \alpha p_i\dfrac{\partial f}{\partial x_i}(x)\right) $ by the Directional Derivative Theorem

$\displaystyle =\alpha \sum_{i=1}^{n}p_i \frac{\partial f}{\partial x_i}(x)$ taking the limit

$\displaystyle =\alpha \dfrac{\partial f}{\partial p}(x)$ by the same theorem.

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  • $\begingroup$ I looks fine though the notation looks a little odd to me. $\endgroup$ – DonAntonio Jul 6 '13 at 23:17
  • $\begingroup$ I am studying from fitzpatrick's advanced calculus book. Here, he used notations so. @DonAntonio Does there is No mistake? Thank you :) $\endgroup$ – user315 Jul 6 '13 at 23:20
  • $\begingroup$ Why do you disturb notations? If there is a mistake, please tell me:( @DonAntonio $\endgroup$ – user315 Jul 6 '13 at 23:23
  • $\begingroup$ I can see no mistake though now it is clearer (perhaps because of the editing): I wasn't sure about that $\;\alpha\;$ , which happens to be a constant. Good, I must have missed that on first reading. $\endgroup$ – DonAntonio Jul 6 '13 at 23:26
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Your proof is incorrect. If you want to use the directional derivative theorem you should not write the definition of partial derivative. Otherwise you have no more a derivative to which apply the theorem.

The proof is much simpler and also holds for functions which are not differentiable but only have the derivative in the direction considered.

If $\alpha=0$ the result is trivial. otherwise just make a change of variables $s=\alpha t$ in the limit defining the partial derivative: $$\frac{\partial f}{\partial \alpha p}(x) = \lim_{t \to 0}\frac {f(x+t\alpha p)-f(x)}{t} = \lim_{s \to 0}\frac{f(x+sp)-f(x)}{s/\alpha} = \alpha \frac{\partial f}{\partial p}(x) $$

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  • $\begingroup$ Is that all? So easy! My solution is totally nonsense. Thank you for helping $\endgroup$ – user315 Jul 6 '13 at 23:32
  • $\begingroup$ I don't think your solution is incorrect, @B11b, since from the start you're carrying on derivation wrt $\,\alpha t\,$ , not $\,t\,$ or $\,\alpha\,$ alone . It just looked odd, since in fact it should have been formally $\;\alpha t\to 0\;$ and not only $\,t\to 0\;$ , though it all comes into its correct place once one realizes that $\,t\to 0\iff \alpha t\to 0\;$ (all the time, $\,\alpha\;$ is a constant ...) $\endgroup$ – DonAntonio Jul 6 '13 at 23:35
  • $\begingroup$ @DonAntonio: the equality $\displaystyle\frac {f(x+\alpha tp)-f(x)}{t} = \sum_{i=1}^{n} \alpha p_i\frac{\partial f}{\partial x_i}(x)$ is incorrect. $\endgroup$ – Emanuele Paolini Jul 6 '13 at 23:39
  • $\begingroup$ Hmm okay @DonAntonio but my solution is too complicated. I would not get successful grade from that if i did this solution in an exam paper. And I cannot adjust my notations just like you said. E.Paolini's solution is so clear. :) $\endgroup$ – user315 Jul 6 '13 at 23:40
  • $\begingroup$ @EmanuelePaolini, not if you're derivating with to the whole $\,\alpha t\;$ . You could as well write $\,(\alpha t)p\;$ ...But I agree this would be confusing $\endgroup$ – DonAntonio Jul 6 '13 at 23:43

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