Answer to Question #170334 in Real Analysis for Prathibha Rose

Question

Let f: R² to R defined by f(x,y) =x²+y² .show that f is differentiable at (x₀,y₀) element. Of R² and find gradient of f at (x₀,y₀).

Answer 1

Given f: R² $\to$ R

such that f(x,y) = x² + y²

We have to show that f is differentiable at (x₀,y₀)

given (x₀,y₀) be an arbitrary point in R² . Since f(x₀,y₀) = x₀² + y₀²

f_x(x₀,y₀) = 2x₀ & f_y(x₀,y₀) = 2y₀

we find that the linear approximation to f at (x₀,y₀) is given by :

L(x, y) = x₀² + y₀² + 2x₀(x − x₀) + 2y₀(y − y₀) = 2x₀x + 2y₀y − x₀² −y₀² .

Therefore,

f(x, y) − L(x, y) = x² + y² − ( 2x₀x + 2y₀y − x₀² −y₀² )

= x² + y² − 2x₀x - 2y₀y + x₀² + y₀² )

= (x − x₀)² + (y − y₀)²

It follows that,

$\lim_{(x,y) \to (x_0,y_0)}$ $\frac{f(x, y) − L(x, y) }{ \sqrt{(x − x_0)^2 + (y − y_0)^2}}$ = $\lim_{(x,y) \to (x_0,y_0)}$ $\frac{(x − x_0)^2 + (y − y_0)^2 }{ \sqrt{(x − x_0)^2 + (y − y_0)^2}}$

= $\lim_{(x,y) \to (x_0,y_0)}$$\sqrt{(x − x_0)^2 + (y − y_0)^2}$

= 0

Since the final function to the right of the limit symbol is continuous at (x₀,y₀). Since the necessary limit is zero, f is differentiable at (x₀,y₀).

Now to find Gradient of f at (x₀,y₀)

Given f(x,y) = x² + y²

( we know that grad(f) = $\nabla f$ = < f_x , f_y > = f_x i + f_y j )

therefore for given f(x,y),

f_x = 2x & f_y = 2y

$\nabla f$ = < 2x, 2y > = 2x i + 2y j

$\nabla f$ at (x₀,y₀) = < 2x₀, 2y₀ > = 2x₀ i + 2y₀ j