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


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

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


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

$\displaystyle = \lim_{t \to 0}\left(\frac {f(x+\alpha tp)-f(x)}{t}\right)$ 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.

  • $\begingroup$ 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

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: $$\frac{\partial f}{\partial \alpha p}(x) = \lim_{t \to 0}\frac {f(x+t\alpha p)-f(x)}{t} = \lim_{s \to 0}\frac{f(x+sp)-f(x)}{s/\alpha} = \alpha \frac{\partial f}{\partial p}(x) $$

  • $\begingroup$ 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

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.