Thread: Linear approximation

Let $\displaystyle f:\mathbb R^2\to\mathbb R$ so that $\displaystyle f(x,y)=(s,t)=\left( x+\dfrac{1}{2}\arctan y,y+\dfrac{1}{2}\arctan x \right)$

Find the linear approximation of $\displaystyle f^{-1}$ on a neighborhood of $\displaystyle (s_0,t_0)=f(0,1).$

Don't worry about the invertibility, that was another question which I solved, but now I need help with this one, I don't get how to solve it.

By the chain rule, we get that $\displaystyle Df(0,1)\circ Df^{-1}(s_0,t_0)=I$.

Okay... but, I don't know how to relate it to find the linear approximation.

If you could help me a bit more.

It's pretty straightforward: Since that relation gives that the derivatives involved are invertible we get $\displaystyle Df^{-1}(s_0,t_0)=Df(0,1)^{-1}$

Oh, now I get, so I just need to compute the inverse but do I need to multiply it by a vector or something?

The linear approximation to g(x, y) at (x_0, y_0) is g(x_0, y_0)+ D_g(x_0, y_0)(x, y) where the last term is the product of the matrix D_g(x_0, y_0) with the vector (x, y).