$f(x,y)=\frac{x^2\sin{y^2}}{x^2+y^4}$ if $(x,y)\neq (0,0)$ and $f(0,0)=0$.

Calculate the partial derivatives of $f$ in $(0,0)$.

For $\frac{\partial f}{\partial x}$ I went ahead and put in $y=0$ since we're regarding $y$ as a constant. However that results in the numerator equaling $0$ so I am inclined to say that $\frac{\partial f}{\partial x}=0$. But if I calculate the partial derivative without evaluating at $(0,0)$ I end up with $({x^2+y^4})^2$ in the denominator. So the denominator would equal $0$ if I plugged in the point we're evaluating at. Hence my thoughts would be the partial derivative doesn't exist at $(0,0)$.

The same goes for $\frac{\partial f}{\partial y}$.

What am I doing wrong?

In the next exercise I have to prove that $f$ is differentiable at $(0,0)$. So that means the partial derivatives have to exist, doesn't it? Where is the flaw in my thinking?

  • $\begingroup$ Consider function $f(x)=\frac xx$ if $x\neq 0$, and $f(x)=1$ if $x=0$. That is differentiable and is the constant 1. Your function have $f(x)\to 0$ as $x\to 0$, so if $f(x)=0$ at $x=0$, then $f(x)$ is differentiable. $\endgroup$
    – user202729
    Apr 19, 2016 at 15:49
  • $\begingroup$ @user202729, how would I determine what the partial derivatives are at $(0,0)$ then? $\endgroup$
    – Tori
    Apr 19, 2016 at 16:47

1 Answer 1


For problems like these, you need to go back to the definition of the partial derivative: $$f_x(0,0) = \lim_{h\to0} {f(h,0)-f(0,0)\over h} = \lim_{h\to 0}\frac0h = 0$$ and similarly for $f_y(0,0)$.


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