A factory releases a toxic pollutant which causes two types of damage to a representative area resident whose utility function is $U(S,H,x)= a \cdot \log(S) + b \cdot \log(H) + c \cdot \log (x)$ where $S$ = health level; $H$ = market price of residential property, a measure of quality of housing, such that $b \cdot \log(H)$ represents the satisfaction of the act of residing in the property. In the first place, the health level is moderately reduced by the greater recurrence of allergy symptoms such as coughing and throat soreness, which is given by the function $S = s_0 \cdot \frac{k}{u}$, $u$ being the amount of the toxic pollutant and $k$ being the amount of health services. The constant $s_0$ is positive. The second effect is the deterioration of the painting and external aspect of surrounding properties, which occurs via the function $H = h(u), \frac{\operatorname{d}\!h}{\operatorname{d}\!u} < 0$. The agent's budget constraint is $p \cdot x + r \cdot k = W$, where $p$ is the price of the consumption good. All relevant markets are competitive.
Argue that $\frac{\partial U(k^*(u), h(u), x^*(u))}{\partial u} = \frac{a}{S(k^*)} \cdot \frac{\partial}{\partial u} \left(\frac{s_0 \cdot k^*(u)}{u}\right) + \frac{b}{H(u)} \cdot \frac{\operatorname{d}}{\operatorname{d}\!u} H(u) + \frac{c}{x^*} \cdot \frac{\partial}{\partial u} x^*(u)< 0$, that is, a large exposition to pollution indeed leads to a loss in welfare assuming that $\frac{\partial}{\partial u} \left(s_0 \cdot \frac{k^*(u)}{u} \right) < 0$ and using the information provided in the question.
What I tried to do is: first we setup the maximization problem: $$ \underset{k,u, x}{\operatorname{max}} \,\, U(k, h(u), x) = a \cdot \log \left[s_0 \cdot \frac{k}{u}\right] + b \cdot \log[h(u)]+ c \cdot \log (x) \\ s.t. \quad p \cdot x + r \cdot k = W $$ which gives us the following first order conditions: $$ \begin{cases} \frac{\partial \mathcal{L}}{\partial k} = 0 \rightarrow a \cdot \frac{\frac{s_0}{u}}{s_0 \cdot \frac{k}{u}} - \lambda \cdot r = 0 \therefore \lambda = \frac{a}{k \cdot r} \\ \frac{\partial \mathcal{L}}{\partial x} = 0 \rightarrow \frac{c}{x} - \lambda \cdot p = 0 \therefore \lambda = \frac{c}{x \cdot p} \\ \frac{\partial \mathcal{L}}{\partial u} = 0 \rightarrow a \cdot \frac{u}{s_0 \cdot k} \cdot \left(-\frac{s_0 \cdot k}{u^2}\right) + b \cdot \frac{1}{H(u)} \cdot H'(u) = 0 \therefore a = b \cdot \frac{H'(u)}{H(u)} \cdot u = b \cdot \epsilon_u \\ \frac{\partial \mathcal{L}}{\partial \lambda} = 0 \rightarrow p \cdot x + r \cdot k = W\end{cases}$$ where $$\epsilon_u = \frac{H'(u)}{H(u)} \cdot u$$ is the elasticity of the price of the property with respect to the pollution level.
Using the first, second and last equations, we arrive in: $$x = \frac{c \cdot W}{p \cdot (a+c)}, \qquad k = \frac{a \cdot W}{r \cdot (a+c)}$$
Now substituting the value we've found for $a$ in these equations, we have: $$x^\star(u) = \frac{c \cdot W}{p} \cdot \left(\frac{1}{b \cdot \epsilon_u + c}\right)$$ and $$k^\star(u) = \frac{b \cdot W}{r} \cdot \left(\frac{\epsilon_u}{b \cdot \epsilon_u + c}\right)$$
Now, taking the derivative of $x^\star(u)$ w.r.t to $u$ and rearranging the expression, we find that $$\frac{\partial x^\star(u)}{\partial u} = \frac{c \cdot W}{p} \cdot \left(-\frac{b \cdot \epsilon_u'}{(b \cdot \epsilon_u + c)^2}\right) = -\frac{b \cdot c \cdot W}{p \cdot (b \cdot \epsilon_u + c)^2} \cdot \epsilon_u'$$ and, therefore, the sign of this expression depends solely on the sign of $\epsilon_u'$.
Considering the information given in the question, we can see that $\frac{\partial U(k^*(u), h(u), x^*(u))}{\partial u} < 0$ will occur if $\frac{\partial}{\partial x} x^*(u) < 0 \iff \epsilon_u' > 0$. With that in mind, let's try to define the signal of that derivative: $$ \begin{aligned} \epsilon_u’ &= \left( \frac{H’’(u) \cdot H(u) - H’(u) \cdot H’(u)}{H(u)^2}\right) \cdot u + \frac{H’(u)}{H(u)} \\ &= \frac{H''(u)}{H(u)} \cdot u - \left[ \frac{H'(u)}{H(u)} \right]^2 \cdot u + \frac{H'(u)}{H(u)} \\ &= \frac{H''(u)}{H(u)} \cdot u + \frac{H'(u)}{H(u)} \cdot \left[ 1 - \frac{H'(u)}{H(u)} \cdot u \right] \end{aligned} $$
Now, since $H(u)$ is always positive and $H'(u)$ is negative, $\frac{H'(u)}{H(u)} < 0$ and $1 - \frac{H'(u)}{H(u)} \cdot u > 0$, which means that $\frac{H'(u)}{H(u)} \cdot \left[ \frac{H'(u)}{H(u)} \cdot u \right] < 0$. If we further suppose that $H(u)$ exhibits decreasing marginal returns in $u$, then $H''(u) < 0$ and thus, $\epsilon_u' < 0$.
Since I found that $\epsilon_u' < 0$, that means $\frac{\partial}{\partial u} x^\star(u) > 0$ and because of that I cannot ascertain whether $\frac{\partial U(k^*(u), h(u), x^*(u))}{\partial u} < 0$ or not. What did I do wrong?
I appreciate any help!
EDIT: Adding the rest of the questions, to see if it helps.
a) Find the marginal impact of the degree of pollution $u$ in the acquired quantities of the consumption goods and the health services.
b.1) Show that the marginal impact of the degree of pollution in the "utility level of equilibrium" is $\frac{\partial U(k^*(u), h(u), x^*(u))}{\partial u} = \frac{a}{S(k^*)} \cdot \frac{\partial}{\partial u} \left(\frac{s_0 \cdot k^*(u)}{u}\right) + \frac{b}{H(u)} \cdot \frac{\operatorname{d}}{\operatorname{d}\!u} H(u) + \frac{c}{x^*} \cdot \frac{\partial}{\partial u}$.
b.2) Argue that $\frac{\partial U(k^*(u), h(u), x^*(u))}{\partial u} = \frac{a}{S(k^*)} \cdot \frac{\partial}{\partial u} \left(\frac{s_0 \cdot k^*(u)}{u}\right) + \frac{b}{H(u)} \cdot \frac{\operatorname{d}}{\operatorname{d}\!u} H(u) + \frac{c}{x^*} \cdot \frac{\partial}{\partial u} x^*(u)< 0$, that is, a large exposition to pollution indeed leads to a loss in welfare assuming that $\frac{\partial}{\partial u} \left(s_0 \cdot \frac{k^*(u)}{u} \right) < 0$ and using the information provided in the question.
