My question concerns the proof of Theorem 2 in §11.3 of PDE Evans:
THEOREM 2 (Riemann invariants and blow-up). Assume $\mathbf{g}$ is smooth, with compact support. Suppose also the genuine nonlinearity condition $$\frac{\partial \lambda_i}{\partial w_j} > 0 \quad \textit{in } \mathbb{R}^2 \quad (i=1,2, \,\, i\not= j) \tag{27}$$ holds. Then the initial-value problem $\text{(1)}$ cannot have a smooth saolution $\mathbf{u}$ existing for all times $t \ge 0$ if $$\textit{either} \quad v_x^1 < 0 \quad \textit{or}\quad v_x^2 < 0 \quad \textit{somewhere on }\mathbb{R} \times \{t=0\}. \tag{28}$$
The context for this question is too involved; the proof is very long and references too many previous equations throughout in the section, so here is the link to §11.3 of PDE Evans (1st edition) (PDF), which contains the above theorem, and its long proof, starting on page 597.
My question:
Why does the second equality of
$$\frac d{dt}((\xi \alpha)^{-1})=\frac{-1}{(\xi \alpha)^2} \frac d{dt}(\xi \alpha) = \frac{\partial \lambda_2}{\partial w_1} \xi^{-1}$$
hold? (This equality in question is seen towards the end of the proof in the PDF file.) NB: I realize that the first inequality results from simply taking the derivative.
My work so far:
Equation $(31)$ in §11.3 says:
$$a_t + \lambda_2 a_x + \frac{\partial \lambda_2}{\partial w_1} a^2 + \left[\frac 1{\lambda_2 - \lambda_1} \frac{\partial \lambda_2}{\partial w_2} (v_t^2+\lambda_2 v_x^2) \right]a=0 \tag{31}$$
where $a := v_x^1$.
Anyways, I integrated and exponentiated, $(31)$, then simplified a bit to obtain $$\exp\left(\int_0^t a_t + \lambda_2 a_x + \frac{\partial \lambda_2}{\partial w_1} a^2 \, ds \right)\xi(t)=1 \tag{$\star$},$$ where
$$\xi(t) := \exp\left(\int_0^t \frac 1{\lambda_2-\lambda_1} \frac{\partial \lambda_2}{\partial w_2} (v_t^2+\lambda_2 v_x^2)(x_1(s),s) \, ds \right). \tag{32}$$
Am I doing this right so far? If so, then my next steps are to solve from $(\star)$ for $\alpha(t) \xi(t)$, then find $\frac d{dt}(\xi \alpha)$, so that I can find $\frac{-1}{(\xi \alpha)^2}\frac d{dt}(\xi \alpha)$ as needed. Note from the textbook that $\alpha(t) := a(x_1(t),t)$ and we assume $\alpha \not=0$.
(Again, for those who did not read Evans but might want to answer this question, please refer to either your textbook copy (if you have one) or the link to the PDF file that I provided above. The notation in Evans becomes quickly confusing without reading first §11.3.)