Let X be an $n\times n$-matrix of the form $X(x)=\left (f _{ij}(x) \right )$, where the functions $f _{ij}:V_{K}^{n}\rightarrow \mathbf{K}$ are analytic, and let $Y$ be in $GL_{n}(\mathbf{K})$. Show that the derivative $(Y^{-1}XY)'$ of the function $Y^{-1}XY:V_{K}^{n}\rightarrow M_{n}( \mathbf{K})$, which takes x to $Y^{-1}X(x)Y$, is equal to $Y^{-1}X'(x)Y$ by using the following definition:
Definition: Let $U$ be an open subset of $\mathbf{K}^{n}$ and let $$f:U\rightarrow \mathbf{K}^{m}$$ be a function.If there exists a linear map $g:\mathbf{K}^{n}\rightarrow \mathbf{K}^{m}$ such that $$\lim_{\left \| h \right \|\rightarrow 0}\frac{\left \| f(x+h)-f(x)-g(h) \right \|}{\left \| h \right \|}=0,$$ where $\left \| h \right \|=max_{i}\left | h_{i} \right |$,we say that f is differentiable at x and $f'(x)=g$.
The derivative of the function $$ g:X \mapsto Y^{-1}XY $$ can be found by noting that $$ g(X+H) - g(X) = Y^{-1}(X+H)Y - Y^{-1}XY = Y^{-1} HY $$ So that the derivative of $g$ can be defined by $$ [g'(X)](H) = Y^{-1} HY $$ By the chain rule, we have $$ D [g(X(x))] = [[Dg](X(x))](X'(x)) = Y^{-1}X'Y $$ as desired.