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Let $f=f(x,y)$ s.t. $\partial _x f$ and $\partial _yf$ exist. If $\alpha $ and $\beta $ are differentiable, does $$\frac{d}{dt}f\big(\alpha (t),\beta (t)\big)=\alpha '(t)\partial _xf\big(\alpha (t),\beta (t)\big)+\beta '(t)\partial _y f\big(\alpha (t),\beta (t)\big) \ \ ?$$ The reason I'm asking that it's that indeed for example $$\lim_{h\to 0}\frac{f(\alpha (t+h),u)-f(\alpha (t),u)}{\alpha (t+h)-\alpha (t)}=\partial _x f(\alpha (t),u),$$

whenever $u$ is fixed. But here, I'm not so sure why $$\lim_{h\to 0}\frac{f(\alpha (t+h),\beta (t+h))-f(\alpha (t),\beta (t+h))}{\alpha (t+h)-\alpha (t)}=\partial _x f(\alpha (t),\beta (t))$$ if there is not a stronger assumption as differentiability.

Walace
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  • It is simply the chain rule, applied to the composition of $t \to (\alpha(t), \beta(t))$ with $f$. – Martin R May 20 '20 at 09:39
  • @MartinR: So it works even if $f$ is not differentiable, but only if the partial derivative exist ? – Walace May 20 '20 at 09:41
  • Probably not – it seems that I missed that part of your question :) – Martin R May 20 '20 at 09:41
  • You are right, it is simply not enough to say that $f_x$ and $f_y$ exist. Differentiability needs to be there to justify various limit interchanges – Ninad Munshi May 20 '20 at 09:47

1 Answers1

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You're right to be suspicious. Take

$$f(x,y) = \begin{cases} x & y=0 \\ y & x=0 \\ 0 & \text{else} \\ \end{cases}$$

Here, $f_x$ and $f_y$ exist at $(0,0)$ and equal $1$. However, if we take $\alpha(t) = \beta(t) = t$, then we get that

$$\frac{d}{dt}f(\alpha(t),\beta(t))\Biggr|_{t=0} = 0 \neq 2 = \alpha'(0)f_x(0,0) + \beta'(0)f_y(0,0)$$

Ninad Munshi
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