Question

f(x) = log2[(x+1)/(x-1)] dins the derivative

Answer 1

find derivative?

Answer 2

yes

Answer 3

I would split the logarithm

f(x) = log_2(x+1) - log_2(x-1)

then just use the basic rule

log_n(x) = x'/xlog(n)

Answer 4

you can use chain rule if you want

Answer 5

show me how to solve it using chain rule

Answer 6

so you would have

f'(x) = 1/(x+1)log(2) - 1/(x-1)log(2)

Answer 7

ok I can show you how to use chain rule but give me some time

Answer 8

ok sure :)

Answer 9

Do you want me to use product rule or quotient rule

Answer 10

let me know now or I'm just going to pick :L

Answer 11

use tthe quotient rule

Answer 12

anything no problem

Answer 13

f(x) = log2[(x+1)/(x-1)]

So to deal with chain rule (and most other rules it can help as well such as product etc..) the best course of action when you are new to derivatives is to split the function into smaller functions

so set l(x) to the outer function and g(x) to the inner function

l(x) = log_2(x)
l'(x) = 1/xlog(2)

g(x) = (x+1)/(x-1)
We need to use quotient rule or product rule to find g'(x)
so split it up into smaller functions

s(x) = (x+1)
s'(x) = 1

d(x) = (x-1)
d'(x) = 1

so now I'm going to apply quotient rule to solve g'(x)

which is,

( s'(x)d(x) - s(x)d'(x) )/(d(x))^2

so,

g'(x) = ( s'(x)d(x) - s(x)d'(x) )/(d(x))^2
g'(x) = ( (x-1) - (x+1) )/(x-1)^2

so now we have,

l(x) = log_2(x)
l'(x) = 1/xlog(2)

g(x) = (x+1)/(x-1)
g'(x) = ( (x-1) - (x+1) )/(x-1)^2

sub that into chain rule to solve for f'(x)

chain rule being l'(g(x))*g'(x)
so we have

f'(x) = l'(g(x))*g'(x)

f'(x) = ( 1/((x+1)/(x-1))log(2) ) * ( (x-1) - (x+1) )/(x-1)^2

Answer 14

eventually you should be able to do this in your head but it can still help to write it out to make sure you do not make mistakes when derivatives get really dirty

Answer 15

thak you :))

Answer 16

I don't understand why most teachers don't teach derivatives this way