# linear approximation of $f(x)$

Let $y=f(x)=(x_1^2+2x_2, x_1x_2-3x_1)$

Is the linear approximation just $f(y)=f(x)+A(y-x)$ whenever $y$ is approximately near $x$?

I know that if I calculate the Jacobian matrix, I can get that $Df=\begin{pmatrix} 2x_1 & 2\\ x_2 & 0 \end{pmatrix}\implies Df(1,1)=\begin{pmatrix} 2 & -2\\ 1 & 0 \end{pmatrix}$.

I'm not sure how to proceed from here.

• Your $Df$ is wrong – user251257 Oct 9 '15 at 7:13

## 1 Answer

A differentiable function $f:\mathbb{R}^2\to\mathbb{R}^2$ can be approximated by it's derivative in the sense that given a point, say, $(1,1)$, you can write:

$$f(1+\delta x,1+\delta y)\approx f(1,1)+Df(1,1)(\delta x,\delta y)$$

where $Df(1,1)$ is a matrix applied to the vector $(\delta x,\delta y)$. So, yes, if $y$ is near $x$ (as $(1+\delta x,1+\delta y)$ is near $(1,1)$), then you can approximate the difference $f(y)-f(x)$ by the derivative evaluated at $x$, applied to the difference $y-x$. That is:

$$f(y)\approx f(x)+Df(x)(y-x)$$

Compare this to the one dimensional case, where the tangent line to the graph of a differentiable function locally approximates the graph, in the sense that:

$$f(y)\approx f(x)+f'(x)(y-x)$$