In multivariable calculus, we say $f: U \subseteq \mathbb{R}^d \to \mathbb{R}^m$ is differentiable at $x_0 \in U \subseteq \mathbb{R}^d $, $U$ open if there exists a linear map $T: \mathbb{R}^d \to \mathbb{R}^m$ such that

$$ \lim_{h \to 0 } \frac{f(x_0 + h) - f(x_0) - T(h)}{||h||} = 0 $$

Equivalently, we say that $f$ is differentiable if $\exists T$ such that

$$ f(x_0 + h) - f(x_0) = T(h) + o(||h||) $$ as $h \to 0 $. Now, if we put $h = tx$. Then if $x \neq 0$, have that $h \to 0 \implies t \to 0^+ $ assuming $t$ to be positive. Then, we have

$$ f(x_0 + xt) - f(x_0) = T(xt) + o(||tx||) \implies \frac{ f(x_0 + xt) - f(x_0)}{t} = \frac{t T(x)}{t} + t \;o(||x||)$$

$$ \therefore \frac{ f(x_0 + xt) - f(x_0)}{t} = T(x) + o(||x||)t$$

In the limit $t \to 0^+$, then we obtain that

$$ T(x) = \lim_{t \to 0^+}\frac{ f(x_0 + xt) - f(x_0)}{t} $$

Is this a correct argument to show that $T$ can be expressed as above? thanks for any feedback.

  • $\begingroup$ your argument looks good, however, beware that if you construct $T$ as you indicate then its not sufficient to show differentiability of $f$. What you write for $T$ is the directional derivative of $f$ at $x_o$ in the $x$-direction. In the case that $f$ is differentiable then certainly the differential reproduces directional derivatives. However, there are examples of functions which have directional derivatives in all directions and yet fail to be differentiable. For example: math.stackexchange.com/questions/372070/… $\endgroup$ – James S. Cook Mar 3 '14 at 20:15

Technically, this definition would essentially yield the Jacobian matrix applied in the direction of $x$.

Note that when you define $h=tx$, then you're assuming $h$ gets small along the path defined by $x$. So essentially this becomes a directional derivative. In fact, this is precisely the definition of a directional derivative.

However, the linear map must apply when $h$ gets small generally, not just along some $x$.

  • $\begingroup$ So, $T(x)$ as defined in my last line does not make sense ? $\endgroup$ – ILoveMath Mar 3 '14 at 20:07
  • $\begingroup$ Correct. What you're defining is the the directional derivative in the direction of $x$, but a function can have directional derivatives in every direction, but not be differentiable at a point. See the example here: en.wikipedia.org/wiki/… $\endgroup$ – Emily Mar 3 '14 at 20:09

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