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Define the variable V (implicitly) by:

[itex]V=b + \int^{rV}_{0} rV dF(s) + \int^{\infty}_{rV} s dF(s)[/itex],

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

Question: Show that [itex]\frac{dV}{db}=\frac{1}{1-rF(rV)}[/itex],

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

[itex]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)[/itex]

Rearrangement yields

[itex]\frac{dV}{db}=\frac{1}{1-rF(rV)+rVf(V)(1-r)}[/itex]

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.