Partial derivative of $(x^3+y^3)^{1/3}$

calculusderivativesmultivariable-calculuspartial derivative

Let $f(x,y) = (x^3+y^3)^{1/3}$. Computing the partial derivative with respect to $x$, we get
$$
\frac{\partial f}{\partial x} = \frac{x^2}{(x^3+y^3)^{2/3}}.
$$
This can't be evaluated at $(0,0)$, and the limit doesn't exist at $(0,0)$ either (approaching the origin along the $x$-axis and the $y$-axis give different values). However, if we use the definition of the partial derivative at $(0,0)$ we get
$$
\frac{\partial f}{\partial x} (0,0) = \lim_{h \to 0} \frac{f(0+h,0) – f(0,0)}{h} = \lim_{h \to 0} \frac{(h^3)^{1/3}}{h} = 1,
$$
so the definition seems to tell us that this partial derivative is $1$. Why does differentiating with respect to $x$ and then taking a limit not give the same answer?

Best Answer

What you have is a function $f$ whose first partial derivative $\frac{\partial f}{\partial x}$ exists everywhere but is not continuous.

And that's just how it goes. There is no general guarantee that the partial derivative of a function must be continuous at any point where it exists.

There are even examples of this kind of behavior in 1 variable calculus. The function $$f(x) = \begin{cases} x^2 \sin(1/x) &\quad\text{if $x \ne 0$} \\ 0 &\quad\text{if $x=0$} \end{cases} $$ is differentiable everywhere, but its derivative is not continuous at $x=0$. Applying the limit formula for derivatives at $x=0$ gives $f'(0)=0$, but for values $x \ne 0$ we get $$f'(x) = 2x \sin(1/x) - \cos(1/x) $$ which has no limit as $x$ approaches zero.

This might give you more insight into some of the theorems of multivariable calculus, specifically this one:

Given a function $f : \mathbb R^2 \to \mathbb R$, if its partial derivatives $\partial f/\partial x$ and $\partial f/\partial y$ exist everywhere and are continuous, then $f$ is a differentiable function.

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