Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Mostly I believe in math. However I have trouble in my economic textbook (which really should be right).

I have the following equation:

$$ u(c,d)=\left(ac^{\frac{1-\gamma}{\theta}}+b d^{\frac{1-\gamma}{\theta}}\right)^{\frac{\theta}{1-\gamma}}$$

where $\theta=(1-\gamma)/(1-1/\psi)$, $a\in (0,1),b\in (0,1), c>0, d>0, \gamma>0$, and $\psi\in(0,\infty)$.

Is the derivative of $u$ wrt. $c$ really $$ u_c(c,d)=ac^{\frac{1-\gamma}{\theta}-1}\left(ac^{\frac{1-\gamma}{\theta}}+b d^{\frac{1-\gamma}{\theta}}\right)^{\frac{\theta}{1-\gamma}-1}=ac^{-1/\psi}u(c,d)^{1/\psi}$$

If so, can someone tell me why? I would write it as

$$ u_c(c,d)=ac^{\frac{1-\gamma}{\theta}-1}\left(ac^{\frac{1-\gamma}{\theta}}+b d^{\frac{1-\gamma}{\theta}}\right)^{\frac{\theta}{1-\gamma}-1}=ac^{-1/\psi}u(c,d)^{-1}$$

Hope to hear from someone, thanks in advance.

share|cite|improve this question
Will it help to write $ \log u = 1/\delta ~\log ~(ac^\delta + bd^\delta)$ where $\delta = (1-\gamma)/\theta$? – Ganesh May 10 '13 at 1:42

I think the following is a simpler way. Let $\delta = \frac{1-\gamma}{\theta}$ ; now take the logarithm of both sides, to get:

$$ \log u(c,d) = \log(ac^\delta + bd^\delta) / \delta $$

Differentiating this partially w.r.t $c$ leads to:

$$ \frac{1}{u} \frac{\partial u} { \partial c} = \frac{1}{\delta (ac^\delta+bd^\delta) } (a\delta c^{\delta -1}) $$.

by the chain rule.

Can you now multiply by $u$ on both sides to get $\frac{\partial u} { \partial c}$?

share|cite|improve this answer

$$\frac{\partial}{\partial c}u(c,d)=\frac{\partial}{\partial c}\left(ac^{\frac{1-\gamma}{\theta}}+b d^{\frac{1-\gamma}{\theta}}\right)^{\frac{\theta}{1-\gamma}}$$ $$\Rightarrow \left(ac^{\frac{1-\gamma}{\theta}}+b d^{\frac{1-\gamma}{\theta}}\right)^{\frac{\theta}{1-\gamma}-1}\frac{\partial}{\partial c}\left(ac^{\frac{1-\gamma}{\theta}}+b d^{\frac{1-\gamma}{\theta}}\right)$$ $$\Rightarrow \left(ac^{\frac{1-\gamma}{\theta}}+b d^{\frac{1-\gamma}{\theta}}\right)^{\frac{\theta}{1-\gamma}-1}ac^{\frac{1-\gamma}{\theta}-1}$$ Since $$u=\left(ac^{\frac{1-\gamma}{\theta}}+b d^{\frac{1-\gamma}{\theta}}\right)^{\frac{\theta}{1-\gamma}}\Rightarrow u^{\frac{1-\gamma}{\theta}}=ac^{\frac{1-\gamma}{\theta}}+b d^{\frac{1-\gamma}{\theta}}$$ And $$u_c=\left(u^{\frac{1-\gamma}{\theta}}\right)^{\frac{\theta}{1-\gamma}-1}ac^{\frac{1-\gamma}{\theta}-1}$$ $$\Rightarrow u_c=u^{\frac{\theta+\gamma-1}{\theta}}ac^{\frac{1-\gamma-\theta}{\theta}}=u^{\frac{\theta+\gamma-1}{\theta}}ac^{-\frac{\gamma+\theta-1}{\theta}}$$ Because $$\theta=(1-\gamma)/(1-1/\psi)\Rightarrow \frac1{\psi}=\frac{\theta+\gamma-1}\theta $$ $$\Rightarrow u_c=u^{\frac 1{\psi}}ac^{-\frac1{\psi}}$$

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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