<p>To close this one: As I wrote in my initial comment, one should consider the derivative with respect to $s$, not $c(s)$. This first means that anything containing $t$ has a zero derivative. So we have  </p>

<p>for the numerator</p>

<p>$$\frac {d\ln(c(s)/c(t))}{ds} = \frac {c'(s)}{c(s)}$$</p>

<p>and for the denominator</p>

<p>$$\frac{d[\ln(U'(c(s))/U'(c(t)))]}{ds} = \frac {U''(c(s))\cdot c'(s)}{U'(c(s))}$$</p>

<p>Forming the quotient of the two we get after simplifying and adding the minus sign the desired</p>

<p>$$\lim_{s \to t}\sigma = -\frac{U'(c(t))}{U''(c(t))c(t)}$$</p>

CC BY-SA 3.0
2016-09-30T20:35:12.717Z
2016-09-30T20:35:12.717Z
Alecos Papadopoulos
<p>The inter temporal elasticity of substitution between two periods of time is $$\sigma=-\frac{d[\ln(c(s)/c(t))]}{d[\ln(U'(c(s))/U'(c(t)))]}$$ I need to prove that $$\lim_{s \to t}\sigma = -\frac{U'(c(t))}{U''(c(t))c(t)}$$
So, I have the following by the definition of total differential:</p>

<p>$$\sigma=-\frac{ \frac{dc(s)}{c(s)}-\frac{dc(t)}{c(t)}}{ \frac{U''(c(s))}{U'(c(s)}dc(s)- \frac{U''(c(t))}{U'(c(t)}dc(t)  }$$  Then if $s$ tends to $t$ we have an indetermination $0/0$ so, we use L'Hôpital Rule. That is my problem, I don't know how to derivate with respect to $c(s)$.</p>

<microeconomics><elasticity><economic-growth>
About intertemporal elasticity of substitution
Marcos Fabian C
# About intertemporal elasticity of substitution

(Asked by: username_0 on 2016-02-25)

The inter temporal elasticity of substitution between two periods of time is $$\sigma=-\frac{d[\ln(c(s)/c(t))]}{d[\ln(U'(c(s))/U'(c(t)))]}$$ I need to prove that $$\lim\_{s \to t}\sigma = -\frac{U'(c(t))}{U''(c(t))c(t)}$$
So, I have the following by the definition of total differential:

$$\sigma=-\frac{ \frac{dc(s)}{c(s)}-\frac{dc(t)}{c(t)}}{ \frac{U''(c(s))}{U'(c(s)}dc(s)- \frac{U''(c(t))}{U'(c(t)}dc(t) }$$ Then if $s$ tends to $t$ we have an indetermination $0/0$ so, we use L'Hôpital Rule. That is my problem, I don't know how to derivate with respect to $c(s)$.

**Tags:** microeconomics, elasticity, economic-growth

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## 1 answers

### Answer by username_1 on 2016-09-30. Score: 1

To close this one: As I wrote in my initial comment, one should consider the derivative with respect to $s$, not $c(s)$. This first means that anything containing $t$ has a zero derivative. So we have

for the numerator

$$\frac {d\ln(c(s)/c(t))}{ds} = \frac {c'(s)}{c(s)}$$

and for the denominator

$$\frac{d[\ln(U'(c(s))/U'(c(t)))]}{ds} = \frac {U''(c(s))\cdot c'(s)}{U'(c(s))}$$

Forming the quotient of the two we get after simplifying and adding the minus sign the desired

$$\lim\_{s \to t}\sigma = -\frac{U'(c(t))}{U''(c(t))c(t)}$$


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