Applying the (multivariable) chain rule in matrix notation

Question

If $y_1,y_2$ are implicitly defined by the relations $$\begin{cases}G_1(x,y_1(x),y_2(x))&=0 \\ G_2(x,y_1(x),y_2(x))&=0 \end{cases}$$ find an expression for $$\frac{dy_1}{dx},\frac{dy_2}{dx}$$

The final expression I got was

$$\bigg(\frac{\partial G_1}{\partial y_1} + \frac{\partial G_2}{\partial y_1}\bigg) \frac{dy_1}{dx} + \bigg(\frac{\partial G_1}{\partial y_2} + \frac{\partial G_2}{\partial y_2}\bigg) \frac{dy_2}{dx} = - \bigg(\frac{\partial G_1}{\partial x} + \frac{\partial G_2}{\partial x}\bigg) $$

But the Answer Key in my textbook gives this answer

$$\begin{bmatrix}\frac{dy_1}{dx} \\ \frac{dy_2}{dx}\end{bmatrix} = (-1)\begin{bmatrix}\frac{\partial G_1}{\partial y_1} & \frac{\partial G_1}{\partial y_2} \\ \frac{\partial G_2}{\partial y_1} & \frac{\partial G_2}{\partial y_2}\end{bmatrix}^{-1} \begin{bmatrix}\frac{\partial G_1}{\partial x} \\ \frac{\partial G_2}{\partial x}\end{bmatrix}$$

What did I do wrong and how can I "fix" the original systems of equations so that I'm applying the Matrix version of the Chain Rule in every step of the differentiation process?

Answer 1

Write $G\colon\Bbb R^3\to\Bbb R^2$, with $G(x,y_1,y_2) = \big(G_1(x,y_1,y_2),G_2(x,y_1,y_2)\big)$. Then, provided the matrix $\begin{bmatrix} \frac{\partial G_1}{\partial y_1} & \frac{\partial G_1}{\partial y_2} \\ \frac{\partial G_2}{\partial y_1} & \frac{\partial G_2}{\partial y_2}\end{bmatrix}(x_0,(y_1)_0,(y_2)_0)$ is nonsingular, the Implicit Function Theorem guarantees that the equation $G(x,y_1,y_2)=0$ — with $G(x_0,(y_1)_0,(y_2)_0))=0$ — will locally define $(y_1,y_2)$ as a smooth function of $x$ near the point $(x_0,(y_1)_0,(y_2)_0)$.

If you write $y=(y_1,y_2)$ as a function of $x$, locally, say, $y=\phi(x)$ for $x\in I$, then from $G(x,\phi(x))=0$ for all $x\in I$ we get $$DG(x,\phi(x))\begin{bmatrix} 1 \\ \phi'(x)\end{bmatrix} = 0,$$ which tells you that $$\frac{\partial G}{\partial x}(x,\phi(x)) + \begin{bmatrix} \frac{\partial G_1}{\partial y_1} & \frac{\partial G_1}{\partial y_2} \\ \frac{\partial G_2}{\partial y_1} & \frac{\partial G_2}{\partial y_2}\end{bmatrix}(x,\phi(x))\begin{bmatrix} \frac{dy_1}{dx}\\ \frac{dy_2}{dx}\end{bmatrix} = 0.$$ Since we assumed the matrix $\begin{bmatrix} \frac{\partial G_1}{\partial y_1} & \frac{\partial G_1}{\partial y_2} \\ \frac{\partial G_2}{\partial y_1} & \frac{\partial G_2}{\partial y_2}\end{bmatrix}$ to be nonsingular (invertible), the result now follows.

Answer 2

You summed up the 2 equations, which lose information. What you should have done is doing the derivative separately on each equation, then you will have 2 equations instead and can write it as a matrix.

Or to do it using multivariable chain rule, you can combine $G_{1},G_{2}$ into a single function $G$ that output a 2-dimensional vector. Then the total derivative of $G$ would be a $2\times 3$ matrix, which after chain rule will get applied to a 3-dimensional vector (where one of the coordinate is just $1$ because you're taking derivative with respect to $x$)