Consider $$F(\theta;q;\beta)=\Phi\left(-\frac{\Phi^{-1}(q)-\Phi^{-1}(\theta)\sqrt{1-\beta^2}}{\beta}\right)$$ where $\theta$ denotes the default rate in the portfolio, and $\Phi(\cdot)$ the cumulative distribution function of an $N(0,1)$ variable.
Show that \begin{align*} f(\theta;q;\beta)& =\frac{\partial F(\theta;q;\beta)}{\partial\theta}\\ & = \frac{\sqrt{1-\beta^2}}{\beta}\cdot e^{-\frac{(\Phi^{-1}(q)-\Phi^{-1}(\theta)\sqrt{1-\beta^2})^2}{2\beta^2}+\frac{\Phi^{-1}(\theta)}{2}}. \end{align*}
Hint: First show that $$\frac{\partial\Phi^{-1}(\theta)}{\partial\theta} = \frac{1}{\frac{\partial}{\partial\theta}\Phi(\Phi^{-1}(\theta))}.$$
Solution So this is my solution. \begin{align*} \frac{\partial\Phi^{-1}(q)}{\partial\theta}& =\frac{\partial}{\partial\theta}\left(\int_{-\infty}^{\Phi^{-1}(\theta)}\frac{1}{\sqrt{2\pi}}\cdot e^{-\frac{1}{2}x^2}dx\right)\\ & \overset{1}{=}\left(\int_{-\infty}^{\Phi^{-1}(\theta)}\frac{\partial}{\partial\theta}\cdot\frac{1}{\sqrt{2\pi}}\cdot e^{-\frac{1}{2}x^2}dx\right)\\ & \overset{2}{=}\left(\int_{-\infty}^{\Phi^{-1}(\theta)}\frac{1}{\sqrt{2\pi}}\cdot e^{-\frac{1}{2}x^2}dx\cdot\frac{1}{\frac{\partial}{\partial\theta}(\Phi^{-1}(\theta))}\right)\\ & \overset{3}{=}\frac{1}{\sqrt{2\pi}}\cdot e^{-\frac{1}{2}(\Phi^{-1}(\theta))^2}\cdot\frac{1}{\frac{\partial}{\partial\theta}(\Phi^{-1}(\theta))}\\ & \overset{4}{=}\frac{\partial}{\partial\theta}\Phi(\Phi^{-1}(\theta))\cdot\frac{1}{\frac{\partial}{\partial\theta}(\Phi^{-1}(\theta))}\\ & \overset{6}{=} \frac{1}{\frac{\partial}{\partial\theta}\Phi(\Phi^{-1}(\theta))} \end{align*} However I think I have made a error somewhere since the equals sign number $6$ is just me knowing (guessing) what the answer is. Can anyone help?
The first fact is: $$\tag{1} \Phi'(x)=\frac{1}{\sqrt{2\pi}}e^{-\frac{x^2}{2}}\,. $$ The formula in your hint is wrong. The correct statement of that hint is (derivative of an inverse function) another fact: $$\tag{2} \frac{\partial \Phi^{-1}(\theta)}{\partial \theta}=\frac{1}{\Phi'(\Phi^{-1}(\theta))}\,. $$ Therefore, $$\tag{3} \frac{\partial \Phi^{-1}(\theta)}{\partial \theta}=\sqrt{2\pi}\,\exp\Big(\frac{(\Phi^{-1}(\theta))^2}{2}\Big)\,. $$ Next, by the chain rule, \begin{align}\tag{4} \frac{\partial F(\theta,q,\beta)}{\partial\theta}=\Phi'\Bigg( -\frac{\Phi^{-1}(q)-\Phi^{-1}(\theta)\sqrt{1-\beta^2}}{\beta}\Bigg)\,\frac{\partial \Phi^{-1}(\theta)}{\partial \theta}\frac{\sqrt{1-\beta^2}}{\beta}\,. \end{align} Can you proceed ?