I have a CDF function of $F(x)=p = \int_{0}^{x} \mu e^{-\mu v}e^{-e^{\beta v}\lambda \alpha} dv$.
Using the theorem of calculus, taking derivative of the CDF with respect to $x$ will result to
$$ \frac{dp}{dx} = \mu e^{-\mu x}e^{-e^{\beta x}\lambda \alpha} \rightarrow Equation.1$$
Second derivative of Eq.1 results to
$$ \frac{d^2p}{dx^2} = (\mu e^{-\mu x}e^{-e^{\beta x}\lambda \alpha})(-\mu-e^{\beta x}\beta \lambda \alpha) $$ $$ \frac{d^2p}{dx^2} = (\frac{dp}{dx})(-\mu-e^{\beta x}\beta \lambda \alpha) \rightarrow Equation.2$$
Using substitution of $u=\frac{dp}{dx}$, we can obtain $\frac{d^2p}{dx^2}=\frac{du}{dx}=\frac{du}{dp}.\frac{dp}{dx}=u \frac{du}{dp}$ which simplify Eq.2 into
$$ u \frac{du}{dp} = (u)(-\mu-e^{\beta x}\beta \lambda \alpha) $$ but we still have $x$ term in the equation. Is there alternative way that I can do so that I can solve the CDF before use the solution to invert it? Thank you in advance.