Definition. Let $A \subset R^{m} ;$ let $f: A \rightarrow R^{n} .$ Suppose $A$ contains a neighborhood of a. Given $\mathbf{u} \in \mathbf{R}^{m}$ with $\mathbf{u} \neq \mathbf{0},$ define
$$
f^{\prime}(\mathbf{a} ; \mathbf{u})=\lim _{t \rightarrow 0} \frac{f(\mathbf{a}+t \mathbf{u})-f(\mathbf{a})}{t}
$$
provided the limit exists. This limit depends both on a and on $\mathbf{u}$; it is called the directional derivative of $f$ at a with respect to the vector $\mathbf{u}$. calculus, one usually requires $\mathbf{u}$ to be a unit vector, but that is not necessary.)
Question: Let $A \subset R^{m} ;$ let $f: A \rightarrow R^{n}$. Show that if $f^{\prime}(a, u)$ exists, then $f^{\prime}(a ; c u)$ exists and equals $c f^{\prime}(\mathrm{a} ; \mathrm{u}).$
I couldn't show it. Can you give me a hint?
Best Answer
Just make the substitution $s=tc.$ (Thanks @Kavi Rama Murthy for the hint)
Let $t\neq 0$ and $c\neq 0.$
Since $f'(a;u)$ exists, we have
$$ f^{\prime}(\mathbf{a} ; \mathbf{u})=\lim _{s \rightarrow 0} \frac{f(\mathbf{a}+s \mathbf{u})-f(\mathbf{a})}{s}=\lim _{t \rightarrow 0} \frac{f(\mathbf{a}+tc \mathbf{u})-f(\mathbf{a})}{tc}=\frac{1} {c} \lim _{t \rightarrow 0} \frac{f(\mathbf{a}+tc \mathbf{u})-f(\mathbf{a})}{t}=\frac {1} {c} f'(a;cu) $$
Hence, we get $cf'(a;u)=f'(a;cu).$