I looked for few proofs online but was looking for alternate, more direct proofs. The one on Khan Academy used the Linear Approximation and one used the chain rule of multivariable functions. Are there any alternate methods to prove it? I'm looking for one which doesn't use much except the basic definition of the partial Derivative.

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    You can't get it from the definition of the partial derivative, as you need more for the formula to hold than just having partial derivatives at the point. – Ted Shifrin Aug 29 at 20:56
  • You'll never understand multivariate calculus if you are refusing to go beyond rote computation of partial derivatives. – Christian Blatter Aug 30 at 7:18
  • Fair enough. I just wanted to see an elementary proof. That doesn't mean I'm not willing to go beyond just the definitions. – Star Platinum ZA WARUDO Aug 30 at 7:34
  • @StarPlatinumZAWARUDO Please recall that if the OP is solved you can evaluate to accept an answer among the given, more details here meta.stackexchange.com/questions/5234/… – gimusi Sep 17 at 20:25
up vote 0 down vote accepted

The property holds for differentiable functions, indeed by definition of differentiability we have that

$$\lim_{\vec h\to \vec 0} \frac{ f(\vec x_0+\vec h)-f(\vec x_0)-\nabla f(\vec x_0)\cdot \vec h}{\| \vec h\|}=0 \iff f(\vec x_0+\vec h)-f(\vec x_0)=\nabla f(\vec x_0)\cdot \vec h+o(\| \vec h\|)$$

and assuming $\vec h = t\,\vec v$ we have

$$\frac{\partial f}{\partial \vec v}(\vec x_0)=\lim_{t\to 0}\frac{f(\vec x_0+t\vec v)-f(\vec x_0)}{t}=\lim_{t\to 0}\frac{\nabla f(\vec x_0)\cdot t\vec v+o(\|t\vec v\|)}{t}=\nabla f(\vec x_0)\cdot \vec v$$

  • What is $o(\| \vec h\|)$ and why does it randomly appear? Is it like an identity element? – Star Platinum ZA WARUDO Aug 30 at 2:46
  • It’s the little-o notation which gives an equivalent statement for the limit and differentiability. It represents a reminders duch that $o(h)/h \to 0$. – gimusi Aug 30 at 5:41

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