Thread: Directional Derivative Proof

1. Directional Derivative Proof

Let $A\subset\Re^m$ ; let $f:A\rightarrow\Re^n$ Show that if $f'(a;u)$ exists, then $f'(a:cu)$ exist and equals $cf'(a;u).$

Some people in my class were trying to say that we know that by theorem

$f'(a:u)=Df(a)*u$

But I disagree since we don't know that Df(a) exist but I don't know how to prove this result. Any ideas would be helpful.

2. Originally Posted by tedii
Let $A\subset\Re^m$ ; let $f:A\rightarrow\Re^n$ Show that if $f'(a;u)$ exists, then $f'(a:cu)$ exist and equals $cf'(a;u).$

Some people in my class were trying to say that we know that by theorem

$f'(a:u)=Df(a)*u$

But I disagree since we don't know that Df(a) exist but I don't know how to prove this result. Any ideas would be helpful.
Does $f'(a;u)$ indicate the derivative of f in the direction of unit vector u? Yes, you are right. The derivative in a specific direction may exist even if the function is not "differentiable".

You can prove it directly from the definition:

The definition of the derivative, in the direction of vector u, is $f'(a;u)= \lim_{h\to 0}\frac{f(a+ hu)- f(a)}{h}$

It follows that $f'(a;cu)= \lim_{h\to 0}\frac{f(a+ hcu)- f(a)}{h}$.

Multiply numerator and denominator by c to get
$\lim_{h\to 0}c\frac{f(a+chu)- f(a)}{ch}$

Now let, k= ch. Since c is a constant, k goes to 0 as h goes to 0 and we can write that
$c\lim_{k\to 0}\frac{f(a+ku)- f(a)}{k}= c f'(a;u)$.

3. Tricky tricky, your right that makes perfect sense now, and I can wrap my head around that way of doing it. Thank you very much.