# Taking a complicated partial derivative

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.

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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 $\frac{\partial u} { \partial c}$?

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

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