Malliavin derivative of a gaussian Let $W$ be an $H$-isonormal Gaussian process and $H$ is a real separable Hilbert Space.
Set $$X=f\big(W(h_1),\ldots, W(h_n)\big) $$ for $f$ an infinite differentiable with their partial derivatives have polynomial growth.  
We define the Malliavin derivative as 
$$DX=\sum_{i=1}^n \partial_i f\big(W(h_1),\ldots, W(h_n)\big)h_i.$$
I would like to compute the malliavin derivation of $X=W(h)$ where $h\in L^2(0,\infty) $. Its $DX=h$ but I don't see how can we prove this.  

Simply apply the definition with $n=1$, $h_1=h$, $f(x)=x$.  Note that $f$ is smooth and $\partial_1 f = f' = 1$ has polynomial growth so the hypotheses are satisfied.  Now your formula says exactly that $DX = h$.
