Second order partial derivative 9

Question

I am following a work where the authors start a production function and get this first order partial derivative (with respect to $ k $):

$$ f'(k)=\frac{a(1+bk)}{k^{1-a}(1+abk)^{a}} $$

And by deriving it again obtain the following second order partial derivative:

$$ f''(k)=a(a-1)k^{a-2}(1+abk)^{-a-1} $$

I managed to get the first one but not the second one. Is there someone who can help me with this? Thank u so much and Happy Christmas!! :)

Answer 1

I think that this is a good problem for logarithmic differentiation.

Consider $$y=\frac{a(1+bk)}{k^{1-a}(1+abk)^{a}}$$ $$\log(y)=\log(a)+\log(1+bk)-(1-a)\log(k)-a\log(1+abk)$$ Differentiate both sides $$\frac{y'}y=\frac {b}{1+bk}-\frac{1-a}k-\frac{a^2b}{1+abk}=\frac{a-1}{k (1+b k) (1+a b k)}$$ Now $$y'=y \times \frac{y'}y=\frac{a(1+bk)}{k^{1-a}(1+abk)^{a}}\times \frac{a-1}{k (1+b k) (1+a b k)}$$

Just simplify.

Merry Xmas

Answer 2

Ugly exercise. You can write $f'(k)=a(1+bk)k^{a-1}(1+abk)^{-a}$. Therefore $$f''(k)=abk^{a-1}(1+abk)^{-a}+a(1+bk)(a-1)k^{a-2}(1+abk)^{-a} +a(1+bk)k^{a-1}(-a)ab(1+abk)^{-a-1} $$ $$ = ak^{a-2}(1+abk)^{-a-1}[ bk(1+abk) + (1+bk)(a-1)(1+abk) + k(1+bk)(-a)ab] = $$ $$ = ak^{a-2}(1+abk)^{-a-1}(bk+ab^2k + a + a^2bk -1 - abk + bka +a^2b^2k^2 -bk -ab^2k^2 - a^2bk - a^2b^2k^2) $$ and after simplifications you get the required results. Happy Christmas!