in my cal 4 class, I was given the implicit function
$$\sin(6x+y)=4x^2+7y$$
where $y$ is a function of $x$ with $y(0)=0$
and asked to solve for $\frac{d^4y}{dx^4}(0)$
I was able to determine that the answer must be $\frac{196}{3}$ using implicit differentiation and solving for $\frac{dy}{dx}(0)$, $\frac{d^2y}{dx^2}(0)$, and $\frac{d^3y}{dx^3}(0)$ as I would have in an earlier class. However, given that the topic we are currently studying involve looking at derivatives as matrices, I assume that there is a way of solving this problem using matrices. Unfortunately, my professor is rather unclear and, based on his instruction, I do not know how I would go about it.
My attempt is as follows $$ \begin{align}&f=\sin(6x+y)-4x^2-7y=0\ \mathrm{and}\ g=\begin{bmatrix}x\\y\\\end{bmatrix}\\&\mathrm{thus}\ h=f\circ g\\&h'=\begin{bmatrix}\frac{\partial f}{\partial x}&\frac{\partial f}{\partial x}\\\end{bmatrix} \begin{bmatrix}\frac{\partial x}{\partial x}\\\frac{\partial y}{\partial x}\\\end{bmatrix}=\begin{bmatrix} 6\cos(6x+y)-8x &\cos(6x+y)-7\\\end{bmatrix}\begin{bmatrix}1\\y'\\\end{bmatrix}\\&\mathrm{therefore}\ h'=6\cos(6x+y)-8x+y'\cos(6x+y)-7=(6+y')\cos(6x+y)-8x-7=0\end{align}$$
From here, we input the initial conditions and solve for $y'(0)=\frac{dy}{dx}(0)$. We then repeat the process instead using $h'$ and $g_1=\begin{bmatrix}1&y'&y''\end{bmatrix}^\mathsf{T}$ with initial conditions $y(0)=0;\ y'(0)$ in order to solve for $y''(0)$ and so on. I realize that this method does not have any benefit over the matrix-free method, except perhaps in terms of organization. That leads me to believe that there is some alteration to this process that I could make in order to improve it.
Does this process make sense? Also, is there some way of improving it?
Thanks!