# Show that f with a tangent plane has directional derivatives in all directions at the point $(0, 0).$

Suppose that the continuous function $$f: \Bbb R^2 \to \Bbb R$$ has a tangent plane at the point $$(0, 0, f(0, 0)).$$ Prove that the function $$f$$ has directional derivatives in all directions at the point $$(0, 0).$$

My attempt: The tangent plane at the point $$(0, 0)$$ has the form $$z=ax+by+f(0,0)$$ with the property $$\lim_{(x,y)\rightarrow(0,0)}\frac{f(x,y)-ax-by-f(0,0)}{\sqrt{x^2+y^2}}=0 \ \ (1).$$

The directional derivative of $$f$$ in the direction $$p=(c,d)$$ at $$(0,0)$$ is $$\frac{\partial f}{\partial p}(0,0)=\lim_{t\rightarrow 0}\frac{f(tc, td)-f(0,0)}{t} \ \ (2).$$

How can I use $$(1)$$ to prove that $$(2)$$ exists?

Your limit $$(1)$$ is valid for any path of approach towards the origin. So choose the path $$y = \frac{d}{c}x$$, and plug that in:

$$0 = \lim_{x \to 0} \frac{f(x,\frac{d}{c}x) - \left(a+ \frac{bd}{c}\right)x - f(0,0)}{\frac{x}{c} \sqrt{c^2+d^2}} = \lim_{x \to 0} \frac{f(x,\frac{d}{c}x)-f(0,0)}{\frac{x}{c}\sqrt{c^2+d^2}} - \frac{ ac + bd }{\sqrt{c^2+d^2}}$$

Move the negative term to the other side, and cancel the $$\sqrt{c^2+d^2}$$ factor:

$$ac+bd = \lim_{x \to 0} \frac{f(x,\frac{d}{c}x) - f(0,0)}{x/c}$$

Just re-parameterize $$t = \frac{x}{c}$$, and this becomes

$$ac + bd = \lim_{t \to 0} \frac{f(ct,dt)-f(0,0)}{t}$$

• where did the denominator $\sqrt{x^2+y^2}$ go? – dxdydz Feb 9 at 4:54
• Oops, you're right...changing it right now... – Nick Feb 10 at 0:38