Showing that $C(\theta)$ is strictly increasing in $\theta$ and the relationship between $X$ and $A$

Question

Given $U''<0<U'$ and $f'>0$, $f'(k)k=\alpha f(k)$ and this equation below

$$\left[ f(k)-f'(k)k \right]\theta U'(C) = 1$$.

How do I show that C is strictly increasing in $\theta$?

If I totally differentiate then I get something like this:

$$\frac{d C}{d\theta} = \frac{1}{\left[ f(k)-f'(k)k \right]\theta U''(C)} < 0.$$

Any idea what I am doing wrong here?

Another similar problem: $U'(X)=A$, where A is a constant.

If I totally differentiate then I get $$\frac{d X}{d A} = \frac{1}{U''(X)} < 0.$$ Is this correct?

Answer 1

Since you only care about $\frac{dC}{d\theta}$, you can rewrite your expression as $$\theta U'(C) = \frac{1}{\left[ f(k)-f'(k)k \right]}.$$ If you consider the LHS as a function $F(C,\theta)=\text{some constant}$, the total differentiation gives $$ d\theta\, U'(C) + \theta U''(C)dC = 0 \implies \frac{dC}{d\theta} = -\frac{U'(C)}{\theta U''(C)}$$ Since $U''<0<U'$, it follows that $\frac{dC}{d\theta} >0$ for $\theta >0$.