Relation between the solution (X,Y) of a Backward Stochastic Diffrential Equation and a Brownian motion W.

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If one have the following BSDE:

$x(t)$+$\int_{t}^1f(s,x(s),y(s))ds$+$\int_{t}^1[g(s,x(s))+y(s)]dW(s)$=X

where $\{W_t, t\in [0,1]\} $ is the standard $k$-dimensional Wiener process defined on $(\Omega, \mathscr{F},P)$, $\{\mathscr{F}_t\}_{t\leq1}$ is its natural filtration and X is given. My question is, Why do both the solution $\{(x(t),y(t)\}$ and $\{W(t)\}$ satisfy:

$y(t)=\frac{d}{dt}\langle x,W\rangle_t-g(t, x(t)),$ $\quad$ $t$ a.e.?

where $\langle x,W\rangle_t$ denotes the joint quadratic variation process between $x$ and $W$.