Is for a curve existence of partial derivatives equivalent to differentiability?

Question

Let $\gamma : I\subset \mathbb{R}\rightarrow \mathbb{R}^n $.

$\gamma'(a)=\lim_{t\rightarrow 0} \frac{\gamma(a+t)-\gamma(a)}{t}$.

For diff. at $a$, we must have $\lim_{t\rightarrow 0} \frac{||\gamma(a+t)-\gamma(a)-\gamma'(a)t||}{|t|}=0$.

$\lim_{t\rightarrow 0} \frac{||\gamma(a+t)-\gamma(a)-\gamma'(a)t||}{|t|}=\lim_{t\rightarrow 0} ||\frac{\gamma(a+t)-\gamma(a)}{t}-\gamma'(a)||=||\lim_{t\rightarrow 0} \frac{\gamma(a+t)-\gamma(a)}{t} -\gamma'(a)||=0$, where for the first equality I used $|s|||x||=||sx||$, and for the second continuity of $||\cdot ||$.

So, existence of partial derivatives for curves implies differentiability, right?

Any help would be appreciated.

Proof that continuous partial derivatives implies differentiability — Winther, Dec 30 '16 at 17:18
@Winther thanks for the comment. I know that theorem. But I'm looking for an answer to a different question. Could you help me? — An old man in the sea., Dec 30 '16 at 17:21
@Boris I'm not sure I understand what you mean. Could you elaborate a bit? thanks ;) — An old man in the sea., Dec 30 '16 at 17:33
For the special case $f : \mathbb{R}^m \to \mathbb{R}^n$ where $m=1$, i.e. a simple vector-valued function, then yes: if all the partial derivatives exists then $f$ is differentiable (almost by definition). — Winther, Dec 30 '16 at 17:34

score 0 · Accepted Answer · answered Dec 30 '16 at 17:47

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Your proof is correct. However, you should note that the trick fails when the domain has more than one dimension because you cannot apply the trick of putting the denominator inside dividing as it will be a vector and not an scalar.

answered Dec 30 '16 at 17:47

Josué Tonelli-Cueto

4,752

Is for a curve existence of partial derivatives equivalent to differentiability?

1 Answers1