# A simple computation using Leibniz's Rule

1. Jan 22, 2012

### jakemf1986

A simple computational question, that I saw in a journal article and have been having trouble getting.

Define the variable V (implicitly) by:

$V=b + \int^{rV}_{0} rV dF(s) + \int^{\infty}_{rV} s dF(s)$,

where r is a constant and F has support on [0,∞).

Question: Show that $\frac{dV}{db}=\frac{1}{1-rF(rV)}$,

My attempted solution: Differentiate with respect to V and use Leibniz's Rule to get

$1=\frac{db}{dV} + r\cdot rVf(V) - 0 + \int^{rV}_0 r dF(s) + 0 - rVf(V) + 0 = \frac{db}{dV} + r^{2} Vf(V) + rF(rV) - rVf(V)$

Rearrangement yields

$\frac{dV}{db}=\frac{1}{1-rF(rV)+rVf(V)(1-r)}$

Notice that my solution has an additional ugly term in the denominator.

Is my solution wrong? Or could perhaps a mistake have been made in the article?

Thank you.

2. Jan 25, 2012

### jakemf1986

bump =)

3. Jan 25, 2012

### chiro

Hey jakemf1986 and welcome to the forums.

I think you have differentiated at least one of the integral terms incorrectly since one of the limits involves V. When this happens you need to use a special kind of chain rule.

Does this help?