# Tutor profile: Rodrigo C.

## Questions

### Subject: Calculus

Find the derivative of $$f(x)=ln(x^\sigma-\sigma)$$

For this, we first find the outter derivative, which we, in turn, multiply by the inner derivative due to the chain rule: $( \frac{f(x)}{x}= \frac{1}{x^\sigma-\sigma} \times \frac{d}{dx}(x^\sigma-\sigma) $) $( \frac{f(x)}{x}= \frac{1}{x^\sigma-\sigma} \times \sigma x^{\sigma-1} $) $( \frac{f(x)}{x}= \frac{\sigma x^{\sigma-1}}{x^\sigma-\sigma} $)

### Subject: Economics

Assume consumers demand two goods, bananas ($$x_1$$) and coconuts ($$x_2$$) such that their average utility can be characterized as: $( U=x^{\alpha}_{1}x^{\beta}_{2} $) And have the following budget constraint: $( m=p_1x_1+p_2x_2 $) Where $$m$$ represents the budget of the representative consumer. Derive their demand functions for coconuts and bananas.

First, we may find the marginal rate of substitution, given by $$\frac{\partial x_2}{\partial {x_1}}$$ or $$\frac{\frac{\partial {U}}{\partial {x_1}}}{\frac{\partial {U}}{\partial {x_2}}}$$ In which, $(\frac{\partial {U}}{\partial {x_1}} = \alpha x^{\alpha-1}_{1}x^{\beta}_{2}$) and $(\frac{\partial {U}}{\partial {x_2}} = \beta x^{\alpha}_{1}x^{\beta-1}_{2}$) We arrive at the following MRS: $(\frac{\alpha x^{\alpha-1}_{1}x^{\beta}_{2}}{\beta x^{\alpha}_{1}x^{\beta-1}_{2}} $) Which we may simplify as: $(\frac{\alpha x_2}{\beta x_1} $) Using the identity $$MRS=\frac{p_1}{p_2}$$, we arrive at: $(\beta p_1x_1=\alpha p_2x_2 $) From which we derive the following two identities: $(p_1x_1=\frac{\alpha p_2x_2}{\beta} $) $(p_2x_2=\frac{\beta p_1x_1}{\alpha} $) Which we then are able to plug into the budget constraint to derive the demand for both goods. Starting for $$x_1$$, we obtain: $( m=p_1x_1+\frac{\beta p_1x_1}{\alpha} $) $( m=p_1x_1+\frac{\beta}{\alpha} p_1x_1 $) $( m=(1+\frac{\beta}{\alpha})p_1x_1 $) $( m=(\frac{\alpha+\beta}{\alpha})p_1x_1 $) $( x^{d}_{1}=\frac{\alpha}{\alpha+\beta}\frac{m}{p_1} $) Which is the demand of bananas. Analogously, repeating the same steps we obtain the following demand for coconuts: $( x^{d}_{2}=\frac{\beta}{\alpha+\beta}\frac{m}{p_2} $)

### Subject: Algebra

Solve the following equation for $$x$$: $( x^2-4x=-4 $)

First, we can re-write the equation as: $( x^2-4x+4= 0$) Which, in turn, we can factorize as: $( (x-2)^2= 0$) For which it is only possible to achieve a zero-sum solution given: $( x-2= 0$) From which we arrive at: $( x= 0$)