proving f'(x)/f(x)= 1/(x-a)

Question

I couldn't understand and sole this question. please help 

 

Define f(x):= (x-a1)(x-a2)(x-a3)... (x-at), where an are constants, 1≤n≤t , where t is some finite natural number. Show that

f'(x)/f(x)= (1/x-a1)+(1/(x-a2)+...+(1/x-at)

Answer 1

Zuhair,

 

You need to remember two things to prove this:

First, that the logarithm of a product is the sum of the individual logarithms.  That is ln(ab) = ln(a) + ln(b). And this generalizes to any finite number of factors.

Second, that the derivative of the logarithm of a function is as follows:  d{ln[f(x)]}/dx = f'(x) / f(x), where f'(x) is the first derivative of f(x) wrt to the variable x.

 

For clarity, I should also redefine the function as f(x,t) having 2 arguments: the variable x and the constant t which says how many factors it has.

 

Therefore,  f(x,t) = (x - a₁)(x - a₂)(x - a₃) ... (x - a_t), where the an are constants 1<= n <= t

 

Take the natural logarithm (ln) of both sides which maintains the equality

ln[f(x,t)] = ln(x - a₁) + ln(x - a₂) + ln(x - a₃) + ... + ln(x - a_t)

 

Now take the derivative of both side wrt the variable x which also maintains the equality

f'(x,t) / f(x,t)  =  1 / (x -a₁)  + 1 / (x - a₂) + 1/ (x - a₃) + ... + 1 / (x-a_t)

 

That finishes the proof.