I have the following derivative: $$f'(k)=\frac{a(1+bk)}{k^{1-a}(1+abk)^{^{a}}}$$ and have to compute the partial derivative with respect to b, possibly obtaining the following result:
$$\frac{\partial f'(k)}{\partial b}=\frac{ak^{a}(1-a)(1+a+abk)}{(1+abk)^{a+1}}$$
I've tried to do the exercise but I cannot get closer to the desired result than this:
$$\frac{\partial f'{(k)}}{\partial b}=ak^{a}(1+abk)^{-a}\left [ 1-\frac{a(a+abk)}{(1+abk)} \right ]$$
Thank you so much and Happy Christmas! :)
Your answer is correct. It's just a matter of simplifications. \begin{align} \frac{\partial f'(k)}{\partial b} & = \frac{ak^{a}(1-a)(1+a+abk)}{(1+abk)^{a+1}}\\ & = \frac{ak^{a}(1+a+abk-a-a^2-a^2bk)}{(1+abk)^{a+1}}\\ & = \frac{ak^{a}(1+abk-a^2-a^2bk)}{(1+abk)^{a+1}}\\ &=\frac{ak^{a}(1+abk)^{-a}(1+abk-a^2-a^2bk)}{(1+abk)}\\ & = ak^{a}(1+abk)^{-a}\left [ 1-\frac{a(a+abk)}{(1+abk)} \right ] \end{align}