Answer to Question #220196 in Calculus for Dhananjay

Question

Let f : R²→R,f(x,y) =2xy/x²+y² When (x,y) ≠(0,0) and f(x,y) =0, otherwise. How to Check whether or not f has a directional derivative at (0, 0) in the direction θ=π/4? Deduce that the function f is not differentiable at the point (0, 0)?

Answer 1

f : R²→R,f(x, y) = $\dfrac{2xy}{x^2 + y^2}$

We will first find out $\nabla$ ( f )

$\nabla$ ( f ) = $\dfrac{2y^3 - 2x^2y}{(x^2 + y^2)^2} i + \dfrac{2x^3 - 2xy^2}{(x^2 + y^2)^2} j$

Directional derivative is given by $\nabla$ ( f ) . $\overrightarrow{u}$

Where $\overrightarrow{u}$ = $cos \dfrac{\pi}{4} i + sin\dfrac{\pi}{4} j$ = $\dfrac{1}{\sqrt{2}} i + \dfrac{1}{\sqrt{2}} j$

$\nabla$ ( f ) . $\overrightarrow{u}$ = $\dfrac{\sqrt{2}y^3 - \sqrt{2}x^2y}{(x^2 + y^2)^2} + \dfrac{\sqrt{2}x^3 - \sqrt{2}xy^2}{(x^2 + y^2)^2}$

Now as x $\rightarrow$ 0 and y $\rightarrow$ 0 $\nabla$ ( f ) . $\overrightarrow{u}$ does not exist. Hence, the directional derivative of f does not exist at x = 0 and y = 0.

f'(x, y) = $\dfrac{∂f}{∂x} + \dfrac{∂f}{∂y}$ = $\dfrac{2y^3 - 2x^2y}{(x^2 + y^2)^2} + \dfrac{2x^3 - 2xy^2}{(x^2 + y^2)^2}$

Now as x $\rightarrow$ 0 and y $\rightarrow$ 0 f'(x, y) does not exist. Hence, the derivative of f does not exist at x = 0 and y = 0.