Question

If a function f of two variables is differentiable, prove that all it's directional derivatives exist and they can be computed by D_uf=del fu

Answer 1

To prove: If f is differentiable at $X_0$ , then $D_{X_0}f(U)$ exists for all $U ∈ R^3, ||U ||= 1.$ Moreover, $D_{X_0}f(U) = f'(X_0) · U = (f_x(X_0), f_y(X_0), f_z(X_0)) · U.$

The previous theorem says that if a function is differentiable then all its directional derivatives

exist and they can be easily computed from the derivative.

(i) In this example we will see that a function is not differentiable at a point but the directional

derivatives in all directions at that point exist.

Define $f : R^2 → R b_y f(x, y) = \dfrac{x^2y}{x^4+y^2}, when (x, y) \neq (0, 0) \text{ and } f(0, 0) = 0.$

This function is not continuous at (0, 0) and hence it is not differentiable at (0, 0).

We will show that the directional derivatives in all directions at (0, 0) exist. Let $U = (u_1, u_2) ∈R^3, ||U || = 1 \text{ and }0 = (0, 0).$ Then

$lim_{t\rightarrow 0}\dfrac{f(0+tU)-f(0)}{t}=lim_{t\rightarrow 0}\dfrac{t^3u_1^2u_2}{t(t^4u_i^4+t^2u_2^2)}=lim_{t\rightarrow 0}\dfrac{u_1^2u_2}{t^2u_1^4+u_2^2}=0,$

$\text{ if } U_2=0\text{ and } \dfrac{u_1^2}{u_2^2},if u_2\neq 0$

Therefore, $D_0f((u_1, 0)) = 0 \text{ and } D_0f((u_1, u_2)) = \dfrac{u_1^2}{u_2^2} \text{ when } u_2\neq 0$

Hence,If a function f of two variables is differentiable, prove that all it's directional derivatives exist and they can be computed by D_uf=del fu.