$g:\mathbb{R}^2\rightarrow \mathbb{R}$ defined by:
$$g(x,y)=\begin{Bmatrix} (5x^3-3x^2y+7y^3)/(2x+y)^2 &if (2x+y)\neq 0 \\0 &if (2x+y)= 0 \end{Bmatrix}$$
Calculate the first and second order partials derivatives $g_{y}(0,0)$ and $g_{yy}(0,0)$ of $g$ with respect to $y$ at $(0,0)$ if they exist.
Here's my effort:
$$\frac{\partial f}{\partial y}|(0,0)=\lim_{k \mapsto 0}\frac{g\left ( 0,0+k \right )-g\left ( 0,0 \right )}{k}=\lim_{k\rightarrow 0}=\frac{\frac{0-0+7k^3}{\left ( 0+k \right )^2}}{k}=7$$
Then:
$$\frac{\partial f}{\partial y}=\frac{(5x^3-3x^2y+7y^3){}'(2x+y)^2-(5x^3-3x^2+7y^3)((2x+y)^2){}'}{(2x+y)^4}$$
$$\frac{\partial^2 }{\partial y^2}|(0,0)=\lim_{k\rightarrow 0}\frac{\frac{\partial }{\partial y}(0,k)-\frac{\partial }{\partial y}(0,0)}{k}=\frac{0+7k^3-0+0}{k^3}-7=\lim_{k\rightarrow 0}=0$$
Am I doing the right thing here?
Thank you.
You have $f$ in a couple places where you mean $g$, and then you drop it completely in the final line, where it is not entirely clear what you mean. However, you get the correct answers, so maybe you did it right even though the way you did it is complicated and poorly written.
The solution can be more clearly written as:
Let $f(y) = g(0, y)$, then $g_y(0, 0) = \left .{df\over dy}\right |_0$ and $g_{yy}(0, 0) =\left .{d^2f\over dy^2}\right |_0$ .
Now $$f(y) = \begin{Bmatrix} (0 - 0 + 7y^3)/(0+y)^2 & (0+y)\neq 0 \\0 & (0+y)= 0 \end{Bmatrix} = 7y.$$ So ${df\over dy} = 7$ and ${d^2f\over dy^2} = 0$.
Therefore $g_y(0,0) = 7$ and $g_{yy}(0,0) = 0$