Gâteaux and Fréchet derivate

Determine the Gâteaux and the Fréchet derivate of $$f\colon\mathbb{R}^2\to\mathbb{R}^2, z=\begin{pmatrix}x\\y\end{pmatrix}\mapsto\begin{pmatrix}\sin x\cosh y\\\cos x\sinh y\end{pmatrix}.$$

I started with the Gâteaux derivate with any direction $h=(h_1,h_2)$:

$$\lim\limits_{t\to 0}\frac{f(z+th)-f(z)}{t}=\lim\limits_{t\to 0}\frac{1}{t}\begin{pmatrix}\sin(x+th_1)\cosh(y+th_2)-\sin(x)\cosh(y)\\\cos(x+th_1)\sinh(y+th_2)-\cos(x)\sinh(y)\end{pmatrix}$$ This is the differential quotient, isn't it? Is the Gâteaux derivate therefore given by $$Df(x)[h]=\begin{pmatrix}(\sin(x)\cosh(y))'\\(\cos(x)\sinh(y))'\end{pmatrix}?$$

Let me determine the Fréchet derivate later.

No. You didn't differentiate correctly. For example, you need to differentiate $t \mapsto \sin(x+th_1)\cosh(y+th_2)$ with respect to $t$, not $x$, so that terms with $h_1, h_2$ appear.
If a function has a Fréchet derivative $Df(x)$, then the Gâteaux derivative can be easily computed using the chain rule. Let $\phi_{x,h}(t) = x+t h$, and consider the function $f \circ \phi_{x,h}$. The Gâteaux derivative is then $df(x,h) = (f \circ \phi_{x,h})'(0)$, and it follows from the chain rule that $df(x,h) = Df(x)h$.
In this example, $Df(x) = \begin{bmatrix} \cos x \cosh y & \sin x \sinh y \\ -\sin x \sinh y & \cos x \cosh y \end{bmatrix}$, hence $df(x,h) = \begin{bmatrix} h_1 \cos x \cosh y + h_2 \sin x \sinh y \\ -h_1 \sin x \sinh y + h_2 \cos x \cosh y \end{bmatrix}$, where $h = \binom{h_1 } {h_2}$.
• That is how the Gâteaux derivative is defined. See your first limit in the question, you are taking the limit with respect to $t$. Basically you are turning a higher dimensional function into a one dimensional one. – copper.hat Feb 12 '13 at 17:19