I want to calculate the derivative of the normal loss function. This is what I have:
I know that \begin{equation} \mathcal{L}(x) = \phi(x) - x \left(1-\Phi(x)\right) \end{equation} I also computed that \begin{equation} \phi(x) = \frac{1}{\sqrt{2\pi} } \cdot exp \left( -\frac{x^2}{2} \right) \Longrightarrow \frac{\partial \phi(x)}{\partial x} = \frac{1}{\sqrt{2\pi} } \cdot (- x) \cdot exp \left( -\frac{x^2}{2} \right) = -x \phi(x) \end{equation} and \begin{equation} \frac{\partial \Phi(x)}{\partial x} = \phi(x) \end{equation} I can compute then $\mathcal{L}(x)$ as such: \begin{align*} \frac{\partial \mathcal{L}(x)}{\partial x} & = \frac{\partial \phi(x)}{\partial x} - \frac{\partial \left(x \left(1-\Phi(x)\right)\right)}{\partial x} \\ &= \frac{\partial \phi(x)}{\partial x} - \frac{\partial x}{\partial x} \big(1-\Phi(x)\big) - x \frac{ \partial \left(1-\Phi(x)\right)}{\partial x} \\ & = -x\phi(x) - 1 + \Phi(x) + x\phi(x)\\ &= \Phi(x) - 1 \end{align*}
Is this correct?
Your answer is correct. You can also write $L(x)=\phi(x)-x+x\Phi (x)$ which gives $L'(x)=-x\phi(a)-1+(x\phi(x)+\Phi (x))=\Phi (x)-1$.