# Partial Derivatives and Multivariable Calculus

For the function, $$f = \begin {cases} \frac{sin(x^3)}{x^2+y^2}, \text{if } (x,y) \neq (0,0) \\ 0, \text{if } (x,y) = (0,0) \end {cases}$$ compute $$\frac{\partial f}{\partial x}(0,0)$$ and show that $$\frac{\partial f}{\partial x}$$ is not continuous at $$(0,0)$$.

Now I have already calculated the partial derivative if $$(x,y)\neq(0,0)$$ to be $$\frac{\partial}{\partial x}\left(\frac{\sin{\left(x^{3} \right)}}{x^{2} + y^{2}}\right)=\frac{x \left(3 x \left(x^{2} + y^{2}\right) \cos{\left(x^{3} \right)} - 2 \sin{\left(x^{3} \right)}\right)}{\left(x^{2} + y^{2}\right)^{2}}$$ but would the partial derivative at $$(0,0)$$ be $$0$$?

And if so then how can I use it prove discontinuity at $$(0,0)$$ since I have not yet been taught how to calculate limits using $$\epsilon \ and \ \delta$$?

Any help would be appreciated!

You have$$\frac{\partial f}{\partial x}(0,0)=\lim_{h\to0}\frac{f(h,0)-f(0,0)}h=\lim_{h\to0}\frac{\sin(h^3)}{h^3}=1.$$On the other hand, note that, if $$y\ne0$$,$$\frac{\partial f}{\partial x}(2y,y)=\frac{12}{5} \cos \left(8 y^3\right)-\frac{4 \sin\left(8 y^3\right)}{25 y^3}$$and that therefore$$\lim_{y\to0}\frac{\partial f}{\partial x}(2y,y)=\frac{28}{25}\ne1.$$