Differentiating Integral Exponents?

In summary: And we are differentiating with respect to T because that is the variable we are interested in finding the value of at a specific time t.
  • #1
rwinston
36
0
Hi

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

If we rearrange and differentiate

[tex]
Z(t;T) = e^{-\int_{t}^{\tau}r(\tau)d\tau}
[/tex]

We get

[tex]
r(T) = -\frac{\partial}{\partial{T}}(\log{Z(t;T)})
[/tex]

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

Thanks!
 
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  • #2
rwinston said:
My question is: how do we differentiate the exp(-int()) portion? How can we simplify the integral as an exponent?


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

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

You might also try to first solve your equation Z(t;T) for the integral and then taking the derivative.
 
  • #3
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
[tex]\int_t^\tau r(\tau) d\tau[/tex]
so you have the variable of integration in the limits of integration.

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

If that is the case you need to use the chain rule, to get
[tex]\frac{dZ}{dt}= (\int_t^T r(t,\tau)d\tau)e^{\int_t^T r(t,\tau)\dtau}[/tex]
and Laplace's rule:
[tex]\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
HallsofIvy said:
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
[tex]\int_t^\tau r(\tau) d\tau[/tex]
so you have the variable of integration in the limits of integration.

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

Thanks guys - yes, sorry, I had a typo in the integral - it should have been T as the upper limit of integration.
 
  • #5
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}}
[/tex]
[tex]
-\log{Z(t;T)}=\int_{t}^{T}{r(\tau) d\tau}
[/tex]
[tex]
-\frac{\partial}{\partial{T}}=r(T)-r(t)
[/tex]

r(t)=0, so:

[tex]
r(T) = -\frac{\partial}{\partial{T}}\log{Z(t;T)}
[/tex]

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:
  • #6
If I may ask, why is r(t)=0?
 
  • #7
nm got it.
 

What is an integral exponent?

An integral exponent is a whole number that is used as the power in a mathematical expression. It is also known as a natural or counting number exponent. For example, in the expression 2^3, 3 is the integral exponent.

How do you differentiate integral exponents?

To differentiate integral exponents, you use the power rule of differentiation. This rule states that when a variable x is raised to a power n, the derivative is equal to n times x to the power of n-1. For example, the derivative of x^3 is 3x^2.

What is the difference between integral and fractional exponents?

Integral exponents are whole numbers, while fractional exponents are numbers that are not whole. Fractional exponents represent roots, such as the square root or cube root, while integral exponents represent repeated multiplication.

Can integral exponents be negative?

Yes, integral exponents can be negative. A negative integral exponent indicates that the base is in the denominator of the expression. For example, 2^-3 is equivalent to 1/(2^3) or 1/8.

Why are integral exponents important in mathematics?

Integral exponents are important in mathematics because they are used to represent repeated multiplication, which is a fundamental operation in many mathematical concepts. They also help in solving equations, finding derivatives, and understanding the relationships between different mathematical expressions.

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