given function $f:\mathbb{R^2} \to \mathbb{R^2}$ $f(x,y)=(x^3y^2-y,xy^3-x)$ how do we calculate linear approximation to $(fof)(1+h,1+k)$ for $h,k$ near $0$
i got this $f(x_o+ \triangle x,y_o+\triangle y)\approx f(x_o,y_o)+f_x(x_o,y_o)\triangle x+f_y(x_o,y_o)\triangle y$ can i use this?How should i use ? Here i need partial derivative.how can i compute partial derivative for $fof$
thanks in advance
The formula you've presented is appropriate when we have a scalar-valued function, i.e. a function $f : \Bbb{R}^2 \to \Bbb{R}$. In this case, $f$ maps from $\Bbb{R}^2$ to $\Bbb{R}^2$. That said, the formula you presented will be needed!
Suppose we have a function $f : \Bbb{R}^2 \to \Bbb{R}^2$. Then $f$ will map a point $(x, y)$ to another point in $\Bbb{R}^2$, both coordinates of which will depend on $x$ and $y$. That is, $f(x, y)$ will take the form $(u(x, y), v(x, y))$. To linearise $f$, we simply linearise both coordinates, using the given formula. That is, the linearisation of $f$ at a point $(x_0, y_0)$ is given by \begin{align*} &L(x_0 + \Delta x, y_0 + \Delta y) \\ = \, &(u(x_0, y_0) + u_x(x_0, y_0)\Delta x + u_y(x_0, y_0) \Delta y, v(x_0, y_0) + v_x(x_0, y_0)\Delta x + v_y(x_0, y_0) \Delta y). \end{align*} If we write this with column vectors and matrices, it turns into something a bit neater: \begin{align*} L\begin{bmatrix}x_0 + \Delta x \\ y_0 + \Delta y\end{bmatrix} &= \begin{bmatrix} u(x_0, y_0) + u_x(x_0, y_0)\Delta x + u_y(x_0, y_0) \Delta y \\ v(x_0, y_0) + v_x(x_0, y_0)\Delta x + v_y(x_0, y_0) \Delta y \end{bmatrix} \\ &= \begin{bmatrix} u(x_0, y_0) \\ v(x_0, y_0) \end{bmatrix} + \begin{bmatrix} u_x(x_0, y_0)\Delta x + u_y(x_0, y_0) \Delta y \\ v_x(x_0, y_0)\Delta x + v_y(x_0, y_0) \Delta y \end{bmatrix} \\ &= \begin{bmatrix} u(x_0, y_0) \\ v(x_0, y_0) \end{bmatrix} + \begin{bmatrix} u_x(x_0, y_0) & u_y(x_0, y_0) \\ v_x(x_0, y_0) & v_y(x_0, y_0) \end{bmatrix} \begin{bmatrix} \Delta x \\ \Delta y \end{bmatrix} \\ &= f\begin{bmatrix} x_0 \\ y_0\end{bmatrix} + \begin{bmatrix} u_x(x_0, y_0) & u_y(x_0, y_0) \\ v_x(x_0, y_0) & v_y(x_0, y_0) \end{bmatrix} \begin{bmatrix} \Delta x \\ \Delta y \end{bmatrix}. \end{align*} If we call $\vec{x}_0 = \begin{bmatrix} x_0 \\ y_0\end{bmatrix}$ and $\delta\vec{x} = \begin{bmatrix} \Delta x \\ \Delta y \end{bmatrix}$, then this becomes: $$L\left(\vec{x_0} + \Delta \vec{x}\right) = f(\vec{x_0}) + \begin{bmatrix} u_x(\vec{x}_0) & u_y(\vec{x}_0) \\ v_x(\vec{x}_0) & v_y(\vec{x}_0) \end{bmatrix} \Delta \vec{x}.$$ This is the corresponding formula for maps from $\Bbb{R}^2$ to $\Bbb{R}^2$.
So, how do we do this for $f \circ f$? One simple but dirty way would be to simply compute $f \circ f$ (i.e. find out what $f \circ f$ does to an arbitrary point $(x, y)$ explicitly in terms of $x$ and $y$) and then apply the above formula. A slightly more elegant solution is to use the chain rule (which comes in many forms). Essentially, the chain rule can be stated like so:
So, if $L_1$ is the above linearisation, then define $L_2$ to be the linearisation at the point $f(\vec{x}_0)$: $$L_2(f(\vec{x}_0) + \Delta \vec{x}) = f(f(\vec{x_0})) + \begin{bmatrix} u_x(f(\vec{x}_0)) & u_y(f(\vec{x}_0)) \\ v_x(f(\vec{x}_0)) & v_y(f(\vec{x}_0)) \end{bmatrix} \Delta \vec{x}.$$
Let's compose them with an example. I won't use your current example, but I'll use a similarly difficult example: $$f(x, y) = (x^2 y - y^3, x^4 y^2 - 2x).$$ Then $u(x, y) = x^2 y - y^3$ and $v(x, y) = x^4 y^2 - 2x$, so computing partial derivatives, $$\begin{bmatrix} u_x(x, y) & u_y(x, y) \\ v_x(x, y) & v_y(x, y)\end{bmatrix} = \begin{bmatrix} 2xy & x^2 - 3y^2 \\ 4x^3 y^2 - 2 & 2x^4 y \end{bmatrix}.$$ Let's linearise $f \circ f$ at $\vec{x_0} = \begin{bmatrix} 1 \\ 0\end{bmatrix}$. Note that \begin{align*} f(\vec{x}_0) &= f\begin{bmatrix} 1 \\ 0\end{bmatrix} = \begin{bmatrix} 0 \\ -2\end{bmatrix} \\ f(f(\vec{x}_0)) &= f\begin{bmatrix} 0 \\ -2\end{bmatrix} = \begin{bmatrix} 8 \\ 0\end{bmatrix} \\ \begin{bmatrix} u_x(\vec{x}_0) & u_y(\vec{x}_0) \\ v_x(\vec{x}_0) & v_y(\vec{x}_0)\end{bmatrix} &= \begin{bmatrix} 0 & 1 \\ -2 & 0 \end{bmatrix} \\ \begin{bmatrix} u_x(f(\vec{x}_0)) & u_y(f(\vec{x}_0)) \\ v_x(f(\vec{x}_0)) & v_y(f(\vec{x}_0)) \end{bmatrix} &= \begin{bmatrix} 0 & -12 \\ -2 & 0 \end{bmatrix} \end{align*} Our first linearisation becomes $$L_1\left(\begin{bmatrix} 1 \\ 0\end{bmatrix} + \Delta \vec{x}\right) = \begin{bmatrix} 0 \\ -2\end{bmatrix} + \begin{bmatrix} 0 & 1 \\ -2 & 0 \end{bmatrix} \Delta \vec{x},$$ and our second, $$L_2\left(\begin{bmatrix} 0 \\ -2\end{bmatrix} + \Delta \vec{x}\right) = \begin{bmatrix} 8 \\ 0\end{bmatrix} + \begin{bmatrix} 0 & -12 \\ -2 & 0 \end{bmatrix} \Delta \vec{x}.$$ Therefore, if $L$ is the linearisation of $f \circ f$ at $\vec{x_0}$, then \begin{align*} L\left(\begin{bmatrix} 1 \\ 0\end{bmatrix} + \Delta \vec{x}\right) &= L_2\left(L_1\left(\begin{bmatrix} 1 \\ 0\end{bmatrix} + \Delta \vec{x}\right)\right) \\ &= L_2\left(\begin{bmatrix} 0 \\ -2\end{bmatrix} + \begin{bmatrix} 0 & 1 \\ -2 & 0 \end{bmatrix} \Delta \vec{x}\right) \\ &= \begin{bmatrix} 8 \\ 0\end{bmatrix} + \begin{bmatrix} 0 & -12 \\ -2 & 0 \end{bmatrix} \left(\begin{bmatrix} 0 & 1 \\ -2 & 0 \end{bmatrix} \Delta \vec{x}\right) \\ &= \begin{bmatrix} 8 \\ 0\end{bmatrix} + \begin{bmatrix} 24 & 0 \\ 0 & -2 \end{bmatrix} \Delta \vec{x} \end{align*} We can substitute $\Delta x = \begin{bmatrix} h \\ k \end{bmatrix}$, to get $$L\begin{bmatrix} 1 + h \\ k\end{bmatrix} = \begin{bmatrix} 8 \\ 0\end{bmatrix} + \begin{bmatrix} 24 & 0 \\ 0 & -2 \end{bmatrix} \begin{bmatrix} h \\ k \end{bmatrix} = \begin{bmatrix} 8 + 24h \\ -2k \end{bmatrix},$$ hence for small $h, k$, $$(f \circ f)(1 + h, k) \approx (8 + 24h, -2k).$$