directional derivative of a function with discontinuity$f(x,y)= \frac{x^{5}}{\sqrt{9 x^{8}+9(y + 2)^{8}}}$

Question

Okay so this was a long problem which I solved completely. I did all the questions and I'm just looking for someone to check my solutions. (For part $(2)$ I used the method of setting $f(x,-2)$ for the partial derivative with respect to $x$, then setting $f(0,y)$ for partial derivative with respect to $y$. Idk if it's right though) All help is much appreciated, thanks!

screenshot of work from website, transcribed below

Define $f:R^2→R$ by

$f(x,y):= \begin{cases} \frac{x^{5}}{\sqrt{9 x^{8}+9(y + 2)^{8}}}, & > \text{if } (x,y) \neq (0,-2), \\ 0, & \text{if } (x,y) = (0,-2). > \end{cases} $.

$(1)$ Calculate the partial derivatives of $f$ with respect to $x$ and to $y$ assuming $(x,y)≠(0,−2)$.

$ \frac{\partial f}{\partial x} (x,y)= (x^4(x^8+5(y+2)^8))/(3(x^8+(y+2)^8)^{3/2})$

$ \frac{\partial f}{\partial y} (x,y)=\left(-4\left(x^5\right)\left(y+2\right)^7\right)/\left(3\left(x^8+\left(y+2\right)^8\right)^{\frac{3}{2}}\right)$

$(2)$ Calculate $∂f/∂x(0,−2)$, and $∂f/∂y(0,−2)$

$ \frac{\partial f}{\partial x} (0,-2)= 1/3$

$ \frac{\partial f}{\partial y} (0,-2)= 0$

$(3)$ Calculate the directional derivative of $f$ at $(0,−2)$ along a general vector $ v⃗ =(a,b)$

$D_{\vec{v}}(f)(x,y)=1/3a$

$(4)$ Calculate $∇f(0,−2)⋅(a,b) .$

$=1/3a$

$(5)$ By comparing part $(3)$ and $(4)$, what can you conclude?

$a)$ $f$ is $C^1$ and differentiable.

$b)$ $f$ is $C^1$ but not differentiable.

$c)$ $f$ is not $C^1$ but is differentiable.

$d)$ $f$ is not $C^1$ and is not differentiable.

I chose $c)$.

Answer 1

1. Your title is wrong: the function is continuous, despite being piecewise-defined.
2. Parts (1), (2), (4), are correct, and so is the method for (2). (4) is correct because one takes as a definition $\nabla f(0,-2) := (\frac{df}{dx}(0,-2), \ \frac{df}{dy}(0,-2))$.
3. Part (3) is wrong. You cannot use $D_vf(x) = \nabla f(x)\cdot v$ without checking differentiability. Instead, directly compute $$ D_{(a,b)}f(0,-2) := \lim_{t\downarrow 0} \frac{f(0+ta,-2+tb) - f(0,-2)}t.$$ You should see the square root function $\sqrt{\vphantom{f}\dots}$, and also both $a$ and $b$ in the formula.
4. Part (5) is wrong. If $f$ was differentiable (presumably, in the sense of Fréchet, or some equivalent. Check the definition you were given in class, and also the relevant theorems) then we would have $D_vf(x) = \nabla f(x)\cdot v$, and part (3) shows this is wrong.

Some graphs (interactive math3d.org graph link)

$$z=f(x,y)$$

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$$z=\frac{df}{dx}(x,y)$$

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$$z=\frac{df}{dy}(x,y)$$