# Proof of the continuous function having tangent plane has directional derivatives

Suppose that the continuous function $f: \Bbb R^2 \to \Bbb R$ has a tangent plane at the point $(x_0, y_0, f(x_0, y_0))$

Prove that the function $f$ has directional derivatives in all directions at rhe point $(x_0, y_0)$

I guess that I need to use the definitions of the tangent plane and directional derivatives. Hopefully this is right!

But I am telling to you so honestly, I am not good to prove a theorem. Even if somebody gives a hint, I cannot use it to prove this properly. Thus, show me and teach me this proof step by step. Thank you so much for helping:)

• how do you define the tangent plane? What object exists given this data? Commented Jul 7, 2013 at 21:31
• The deifinition of tangent plane $lim_{(x,y) \to (0,0)} \frac{f(x,y)-g(x,y)}{\sqrt {(x-x_0)^2 + (y-y_0)^2}}=0$ Where $g(x,y)=f(x_0,y_0)+f_x(x_0,y_0)(x-x_0)+f_y(x_0,y_0)(y-y_0)$ @JamesS.Cook Commented Jul 7, 2013 at 21:36
• @jamesS.Cook please help me! Commented Jul 7, 2013 at 21:45
• see the answer. (not mine) Commented Jul 7, 2013 at 22:06

The tangent plane at the point $\;u=(x_0,y_0,f(x_0,y_0))\;$ of the function $\;g(x,y,z):=(x,y,z)\;,\;\;z:=f(x,y)\;$ , is just $$\nabla g(x_0,y_0,z_0)\cdot(x-x_0\,,\,y-y_0\,,\,z-f(x_0,y_0))=0$$
This means the gradient of the function $\;f\;$ exists at $\,(x_0,y_0)\;$ and from another question you asked this means there exists the directional derivative of $\;f\;$ in any diretion at the above point.
• Yes I need to prove the existence of the directional derivative of $f$ in any direction at the above point. But how? Commented Jul 7, 2013 at 22:05