0
$\begingroup$

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!! :)

$\endgroup$
  • $\begingroup$ What are the constants in your formula? $\endgroup$ – Dr. Sonnhard Graubner Dec 24 '18 at 12:14
1
$\begingroup$

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

$\endgroup$
1
$\begingroup$

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!

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.