# Question 30439

Apr 7, 2017

See below

#### Explanation:

This is the Implicit Function Theorem taken to the 2nd derivative.

So start with the function: $f \left(L , K \left(L\right)\right) = \text{const}$

The total derivative has 2 partials in it:

$\frac{\mathrm{df}}{\mathrm{dL}} = {f}_{L} + {f}_{K} \setminus K ' = 0$

$\implies K ' = - \frac{{f}_{L}}{{f}_{K}}$

Going again:

$\frac{{d}^{2} f}{{\mathrm{dL}}^{2}} = \frac{d}{\mathrm{dL}} \left(\textcolor{red}{{f}_{L}} + \textcolor{b l u e}{{f}_{K} \setminus K '}\right)$

=color(red)( f_(LL) + f_(LK) \K') + color(blue)( (f_(KL) + f_(KK) \ K' )\ K' + f_K \K'') color(green)(= 0)#

$K ' ' = - \frac{\left(K ' \left({f}_{L K} + {f}_{K L}\right)\right) + {\left(K '\right)}^{2} {f}_{K K} + {f}_{L L}}{f} _ K$

$K ' ' = - \frac{\left(- \frac{{f}_{L}}{{f}_{K}} \left({f}_{L K} + {f}_{K L}\right)\right) + {\left(- \frac{{f}_{L}}{{f}_{K}}\right)}^{2} {f}_{K K} + {f}_{L L}}{f} _ K$

$K ' ' = - \frac{\left({f}_{L L} {f}_{K} - {f}_{L} \left({f}_{L K} + {f}_{K L}\right)\right) + \frac{{f}_{L}^{2} {f}_{K K}}{{f}_{K}}}{{f}_{K}} ^ 2$

Mixed partial are same so:

$K ' ' = - \frac{{f}_{L L} {f}_{K} - 2 {f}_{L} {f}_{L K} + \frac{{f}_{L}^{2} {f}_{K K}}{{f}_{K}}}{{f}_{K}} ^ 2$