How Does Differentiability Imply the Existence of Partial Derivatives?

[cryptoguy]

Homework Statement

A function f(x) : Rⁿ ->R is said to be differentiable at point $$\vec{a}$$ provided that there exists a constant vector $$\vec{c} = (c_1, ... , c_n)$$ such that

$$lim_(\vec{h} -> 0) \frac{f(\vec{a}+\vec{h}) - f(\vec{a}) - \vec{c}*\vec{h}}{||\vec{h}||}$$

Prove that if the multivariable function f(x) (here $$x = x_1, ..., x_n)$$ is differentiable at $$a = (a_1, ..., a_n)$$ then its first order partial derivatives at a exist.

Homework Equations

I know that the partial derivative definition is

The Attempt at a Solution

I've tried a few things but I've encountered a road block of sorts. I think what I have to do is provide the vector c such that the given equation somehow turns into the definition of the first order partial derivative. That means that instead of having f(a1+h1, a2+h2, ...) I need to make all the h's 0 except one, $$h_i$$ so the term would turn into $$f(a1, a2,..., a_i+h, a_{i+1}...)$$ But I'm not sure how to do that with the vector c... I may be way off base...

Thank you for any hints/advice.

 

[Dick]

You want to put $\vec h$ equal to $h \vec {e_i}$. Where e_i is the ith basis vector and take the limit as the real number h approaches 0. Does that help? You can't pick c. That's a given. You can pick a particular form of h.

 

[cryptoguy]

Thank you for the response. What exactly is an "ith basis vector" though?

 

[Dick]

e_i=(0,0,0,...,1,..0,0,0) with the 1 in position i. The same i as in your problem setup.