# Differentiating Integral Exponents?

1. Apr 11, 2008

### rwinston

Hi

I have a question about rearranging the following equation (I saw this in a finance book):

If we rearrange and differentiate

$$Z(t;T) = e^{-\int_{t}^{\tau}r(\tau)d\tau}$$

We get

$$r(T) = -\frac{\partial}{\partial{T}}(\log{Z(t;T)})$$

My question is: how do we differentiate the exp(-int()) portion? How can we simplify the integral as an exponent?

Thanks!

2. Apr 11, 2008

### Pere Callahan

By the fundamental theorem of calculus you have
$$\frac{\partial}{\partial T}\int_t^T{d\tau\,r(\tau)} = r(T)$$

If it is in the exponent you must use the chain rule. As for the second question: I don't understand it.

You might also try to first solve your equation Z(t;T) for the integral and then taking the derivative.

3. Apr 11, 2008

### HallsofIvy

The answer to your second question is: we can't simplify the exponent.

But I have to say I don't understand your notation. Your exponent is
$$\int_t^\tau r(\tau) d\tau$$
so you have the variable of integration in the limits of integration.

Since you only have t and T in Z, did you intend
$$\int_t^T r(\tau) d\tau$$?
And, if so, is T a constant?

If that is the case you need to use the chain rule, to get
$$\frac{dZ}{dt}= (\int_t^T r(t,\tau)d\tau)e^{\int_t^T r(t,\tau)\dtau}$$
and Laplace's rule:
$$\frac{d}{dt}\int_{a(t)}^{b(t)} F(t, \tau)d\tau= \int_{a(t)}^{b(t)} \frac{\partial F}{\partial t} d\tau+ \frac{db}{dt}F(t, b(t))- \frac{da}{dt}F(t,a(t))[/itex] to differentiate the integral. In this case, r is not a function of t so the first term is 0. The upper limit of integration is T, so the second term is 0. a= t so da/dt= 1. The derivative of the integral is just -r(t) which is basically what Pere Callahan was saying. You derivative is [tex]\frac{dZ}{dt}= -r(t)e^{\int_t^T r(\tau)d\tau[/itex] 4. Apr 11, 2008 ### rwinston Thanks guys - yes, sorry, I had a typo in the integral - it should have been T as the upper limit of integration. 5. Apr 11, 2008 ### rwinston Actually I think I was making this more complicated than it needed to be. [tex] Z(t;T) = exp^{-\int_{t}^{T}{r(\tau) d\tau}}$$
$$-\log{Z(t;T)}=\int_{t}^{T}{r(\tau) d\tau}$$
$$-\frac{\partial}{\partial{T}}=r(T)-r(t)$$

r(t)=0, so:

$$r(T) = -\frac{\partial}{\partial{T}}\log{Z(t;T)}$$

Im still not sure why we are differentiating wrt T - I would have expected t. I suspect that this is due to a misunderstanding of the application of the FTOC in this case.

Last edited: Apr 11, 2008
6. Apr 11, 2008

### exk

If I may ask, why is r(t)=0?

7. Apr 11, 2008

nm got it.