Consider a solution $X$ of the stochastic differential equation
$$dX=cXdt+\sqrt{X}dW$$
For $\mathcal{L}(\alpha,t)=E[\exp(-\alpha X_t)]$ I want to show, that
$$\frac{d}{dt}\mathcal{L}(\alpha,t)=(c\alpha-\frac{1}{2}\alpha^2)\frac{d}{d\alpha}\mathcal{L}(\alpha,t)$$
What I did so far is following:
$$\begin{align} (c\alpha-\frac{1}{2}\alpha^2)\frac{d}{d\alpha}\mathcal{L}(\alpha,t) &=(c\alpha-\frac{1}{2}\alpha^2)E[(-X_t)\exp(-\alpha X_t)] =E[(\frac{1}{2}\alpha^2-c\alpha)X_t\exp(-\alpha X_t)]\\ \end{align}$$
By using the function $h(x,y)=\exp(-xy)$ in Ito's lemma, I find
$$\begin{align}\exp(-\alpha X_t) &=h(\alpha,X_t)\\ &=h(\alpha,0)+\int_0^t(-\alpha)\exp(-\alpha X_s)dX_s+\frac{1}{2}\int_0^t\alpha^2\exp(-\alpha X_s)d[X,X]_s\\ &=1+\int_0^t(-\alpha)cX_s\exp(-\alpha X_s)ds+\int_0^t(-\alpha)\sqrt{X_s}\exp(-\alpha X_s)dW_s\\ &\quad\quad+\frac{1}{2}\int_0^t\alpha^2X_s\exp(-\alpha X_s)ds\\ &=1+\int_0^t(\frac{1}{2}\alpha^2-c\alpha)X_s\exp(-\alpha X_s)ds+\int_0^t(-\alpha)\sqrt{X_s}\exp(-\alpha X_s)dW_s \end{align}$$
Now, is it true, that
$$\frac{d}{dt}\mathcal{L}(\alpha,t)=E[\frac{d}{dt}\exp(-\alpha X_t)]=E[(\frac{1}{2}\alpha^2-c\alpha)X_t\exp(-\alpha X_t)]\quad?$$
I find $$\frac{d}{dt}1=0$$
and
$$\frac{d}{dt}\int_0^t(\frac{1}{2}\alpha^2-c\alpha)X_s\exp(-\alpha X_s)ds=(\frac{1}{2}\alpha^2-c\alpha)X_t\exp(-\alpha X_t)$$
but why is
$$\frac{d}{dt}\int_0^t(-\alpha)\sqrt{X_s}\exp(-\alpha X_s)dW_s=0\quad ?$$
Thank you for your attention!
You have
$$\mathbb{E}\int_0^t(-\alpha)\sqrt{X_s}\exp(-\alpha X_s)dW_s=0$$
because this is a Wiener integral. Therefore the derivative is also zero. Does this answer the question?