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Check the following function for partial or total differentiability and calculate the partial/total derivative if it exists.

$f : ℝ^2 → ℝ, f(x,y) = \sqrt[3]{x^2*y^2}$

Attempt:

i dont know if its correct, but I found the definition that the function is differentiable at the point $x_0$ if the following limit exists:

$\qquad\lim\limits_{t → 0}\frac{f(x_0+t)-f(x_0)}{t}$

The partial derivative of $f(x,y)$ with respect to $x$ at $(x_0,y_0)$ is

$\qquad ∂_x f(x_0,y_0) =\lim\limits_{t → 0}\frac{f(x_0+t,y_0)-f(x_0,y_0)}{t}$

if the limit exists. Correspondingly, the partial derivative of $f(x,y)$ with respect to $y$ at the point $(x_0,y_0)$ is equal to:

$\qquad ∂_y f(x_0,y_0) =\lim\limits_{t → 0}\frac{f(x_0,y_0+t)-f(x_0,y_0)}{t}$

The problem with this is that if i want to find the limit $t→0$, i would be dividing by zero. Therefore, the fracture must first be remodeled. I try to demonstrate this for the partial derivative with respect to $x$ and also omit the indices on $x$ and $y$:

$\frac{f(x+t,y)-f(x,y)}{t} = \frac{\sqrt[3]{(x+t)^2*y^2} - \sqrt[3]{x^2*y^2}}{t}$

and now, I don't know how to reformulate that further.

Graham Kemp
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Vek
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  • Hint: You were not asked to find what the limit is, only to verify whether there is one. What tests for convergence or divergence do you know? – Graham Kemp Jun 22 '22 at 10:53
  • @ Graham Kemp do you think that i should first take the partial derivatives with $∂_xf = \frac{2\sqrt[3]{x^2y^2}}{3x}$ and $∂_yf = \frac{2\sqrt[3]{x^2y^2}}{3y}$ and because root functions and fractional rational functions are continuous, it's totally differentiable? – Vek Jun 22 '22 at 12:36
  • You want to determine if the partial derivative exists by ... using the partial derivative? – Graham Kemp Jun 22 '22 at 22:36

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