Derivatives of Log: Find f'(x) for f(x) = log10(x/x-1)

  • Thread starter Thread starter 10min
  • Start date Start date
  • Tags Tags
    Derivatives Log
10min
Messages
7
Reaction score
0

Homework Statement


Question is
f(x) = log10(x/x-1), find f `(x)


Homework Equations


so I used the formula
d/dx loga(u) = 1/uLNa . (d/dx U)


The Attempt at a Solution


(1/(x/x-1)*ln10)*1/(x-1)^2 I used quotent rule for d/dx U part.

editing

Final Answer I got is

1/x(x-1)ln10
or
1/x2-x ln10
Please help
thank you
 
Last edited:
Physics news on Phys.org
it may be easier to use the log property
log(ab)=log(a)+log(b)
log(x/(x-1))=log(x)-log(x-1)
[log10(x/(x-1))]'=[log(x/(x-1))]'/log(10)
=[log(x)-log(x-1)]'/log(10)
=([log(x)]'-[log(x-1)]')/log(10)
 
so many answer is not right
 
10min said:
so many answer is not right

Just a sign error
Quotient rule is
(u/v)'=(u'v-uv')/v^2
so
(x/(x-1))'=(x'(x-1)-x(x-1)')/(x-1)^2
=(1(x-1)-x(1))/(x-1)^2
=-1/(x-1)^2
not
1/(x-1)
also watch your grouping symbols
1/x2-x ln10
is confusing
try
1/[(x^2-x)ln 10]
or

[1/(x^2-x)]/ln 10
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
Back
Top