I am trying to confirm a stated result on my lecture slide.


Given that $A:= \sum_i^n \frac{a_i}{(1+b)^{t_i}}$, where $a_i,b \in \mathbb{R}_+$ and $t_i \in \{t_1,...,t_n\}$ where $0 < t_1 < ... < t_n < \infty$.

Demonstrate that:

$-\frac1A \frac{\partial A}{\partial b} = - \sum_i^n t_i \frac{a_i}{(1+b)^{-t_i}}$

Current progress:

$\frac{\partial A}{\partial b} = - \sum_i^n t_i \frac{a_i}{(1+b)^{t_i + 1}}$


Just from this first step (which could be incorrect), it seems that I can't arrive at what the lecture slide claims.

Note that this is not from a mathematics lecturer so it could be wrong.

Any assistance welcome and appreciated :-)


1 Answer 1


Something is wrong in the lecture statement. Let $n=1,a_1=5,t_1=3.$ And replace $b$ by $x$ so it is a function of $x$ and usual derivatives are the partial. Then $$A=\frac{5}{(1+x)^3}.$$ Now compute $$-\frac{1}{A} A'(x) = \frac{3}{1+x}.$$ But this does not match the right side of the lecure statement, which is $$-\frac{t_1 a_1}{(1+x)^{-t_1}}=-\frac{15}{(1+x)^{-3}}.$$ It looks too different for the lecture statement to be right, even if in you question there was a simple sign error somewhere.

By the way, your first step looks right.


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