Partial derivative in x of $(x^2+y^2)^{\frac{2}{3}}$

Question

Let $f(x,y)=(x^2+y^2)^{\frac{2}{3}}$. Show that:

I did:

$$f'_x (x,y) = \frac{2}{3}(x^2+y^2)^{-\frac{1}{3}}*2x = \frac{4x}{3(x^2+y^2)^{\frac{1}{3}}}$$ At $(0;0):$

$$\lim_{t\rightarrow 0}\frac{f(t;0)-f(0;0)}{t} = \lim_{t\rightarrow0}\frac{(t^2+0)^{2/3}-0}{t} = \\ \lim_{t\rightarrow 0}\frac{t^{4/3}}{t} = \lim_{t\rightarrow0}t^{1/3} = 0$$

Is this correct?

Answer 1

Yes. Your answer is right. Good job!