Question

show that the derivative of [1/(2a)]ln[(a+x)/(a-x)] is 1/[(a^2)-(x^2)]

Answer 1

I got \[\frac{ x-a }{ (a+x)^3}\]

Answer 2

Write the original with Latex?

Answer 3

\[\frac{ 1 }{ 2a } \ln (\frac{ a+x }{ a-x })\]

Answer 4

and the derivative is supposed to be \[\frac{ 1 }{ a^2 - x^2 }\]

Answer 5

How did you approach it?

Answer 6

so my first line is
\[\frac{ 1 }{ 2a }(\frac{ a-x }{ a+x })[\frac{ (-1)(a+x)-(a-x)}{ (a+x)^2 }]\]

Answer 7

I think this might be a product and chain rule kind of problem..

Answer 8

Sorry I thought that's what I was doing, is it right so far?

Answer 9

I don't know if this will make a difference but did you differentiate the 1/2a?

Answer 10

So I thought because a is a constant it's derivative would be 0

Answer 11

Does it state that a is a constant?

Answer 12

Uh my teacher said it was so in a way, yes

Answer 13

Hold on:
\[\LARGE f(x)=\frac{ 1 }{ 2a } \ln (\frac{ a+x }{ a-x })\]
\[\LARGE f'(x)=\frac{ 1 }{ 2a } *(\ln (\frac{ a+x }{ a-x }))'* ((\frac{ a+x }{ a-x }))'+ (\frac{1}{2a})'*\ln (\frac{ a+x }{ a-x })\]

Answer 14

\[\large f'(x)=\frac{ 1 }{ 2a } *(\ln (\frac{ a+x }{ a-x }))'* ((\frac{ a+x }{ a-x }))'+ \ln (\frac{ a+x }{ a-x })*(\frac{1}{2a})'\]

Answer 15

use this
(ln(a+x)-ln(a-x))\2a

Answer 16

I'm dumb, idk.. wanna take over? >.,

Answer 17

*>.<

Answer 18

those are some big arse font
the whole thing doesn't even fit on the screen

Answer 19

break it down into product

Answer 20

@ikram002p can you explain how you got (ln(a+x)-ln(a-x))\2a I'm not sure what's going on there

Answer 21

assign 1/2a as your u
assign ln blahblah into v

Answer 22

you can use any rule you want from that point on

Answer 23

^Assuming they learned sub

Answer 24

ok
ur equation is
\(f(x)=\frac{1}{2a}\ln{\frac{a+x}{a-x}}\)

Answer 25

ln can be "factored" into identity log

Answer 26

\(f(x)=\frac{1}{2a}\ln{\frac{a+x}{x-x}}=\frac{1}{2a}(\ln(a+x)-ln(a-x))\)

Answer 27

ln(a+x) - ln(a-x)

Answer 28

My gosh, those log rules ._.

Answer 29

@damoss They got that from log rules..
http://www.purplemath.com/modules/logrules.htm

Answer 30

hm okay i know log rules, but how come I can't solve it how I would a regular derivative with product rule etc., is my first line of derivative wrong then?

Answer 31

you know how to check if your first order differentiation is correct?

Answer 32

no, there's a way to do that?

Answer 33

take the anti-derivative

Answer 34

anti-derivative...? We did not learn this :c

Answer 35

so
\(f'(x)=\frac{1}{2a}(\frac{a}{a+x}-\frac{1}{a-x})\)

Answer 36

it's reversing back what you did
I am sure you learned inverse

Answer 37

one sec !
a is constant or variable ?

Answer 38

constant!

Answer 39

ohhh geeshhhhh it is a constant!

Answer 40

@nincompoop its constand we dnt need ODE

Answer 41

so we reached f'(x) part
just simplify

Answer 42

sry i made a typo
\(\Huge f'(x)=\frac{1}{2a}(\frac{1}{a+x}-\frac{1}{a-x})\)

Answer 43

\(\Huge f'(x)=\frac{1}{2a}(\frac{a-x}{a^2+x^2}-\frac{a+x}{a^2-x^2})\)

Answer 44

SMH big arse fonts

Answer 45

?

Answer 46

trying to follow along...

Answer 47

\( \Huge f'(x)=\frac{1}{2a}(\frac{-2x}{a^2-x^2})\)

Answer 48

@damoss tell me wich part u cant follow ?

Answer 49

which*

Answer 50

ohk sry ill do it all again i made many typo

Answer 51

no no it's okay i'm following now

Answer 52

A LOT of typos
maybe because your fonts are so ..... nvm

Answer 53

so I've now got \[\frac{ x }{ a(a^2-x^2) }\]

Answer 54

\(f(x)=\frac{ 1 }{ 2a } \ln (\frac{ a+x }{ a-x })\)
\(f'(x)=\frac{ 1 }{ 2a }(\frac{1}{a+x}-\frac{1}{a-x})\)
\(f'(x)=\frac{ 1 }{ 2a }(\frac{(a-x)}{(a+x)(a-x)}-\frac{(a+x)}{(a-x)(a+x)})\)
mmm is it good now @nincompoop :D
nw @damoss check ur qs again i guess it should be
\(f(x)=\frac{ 1 }{ 2a } \ln (\frac{ a-x }{ a+x })\)
to get \(\frac{1}{a^2-x^2}\)

Answer 55

Ahaha @nincompoop the grammar police.
*sorry, I'm just lurking here since I have the same question.

Answer 56

now I have to zoom that size of an ant font

Answer 57

can you simplify your ln result? laughing out loud that thing is annoying to read

Answer 58

ugh

Answer 59

im doing it for
@damoss not for u @nincompoop :P
so any way untill now @damoss got the same result to me
but , we need to check cuz the orgenal qs seems wrong

Answer 60

okay trying to simplify \[\frac{1 }{ 2a } (\frac{ -2x }{ a^2 - x^2 })\]

Answer 61

\[\frac{ x }{ a(a^2 - x^2) }\]

Answer 62

it wont simplify more than this :)

Answer 63

what is going on T__T i don't know think the answer is wrong

Answer 64

when \(f(x) =\frac{ 1 }{ 2a } \ln (\frac{ a+x }{ a-x })\)
then \(\frac{1 }{ 2a } (\frac{ -2x }{ a^2 - x^2 })\)

when \(f(x) =\frac{ 1 }{ 2a } \ln (\frac{ a-x }{ a+x })\)
then \( f'(x)=\frac{ 1 }{ a^2 - x^2 } \)

Answer 65

so ?

Answer 66

use the fact that ln( (a-x) / (a+x) = ln(a-x) - ln(a+x). Now it's much easier to do the derivative

Answer 67

@sourwing :)
we did the same
but it wont be simplify to
1/a^2-x^2

Answer 68

@ganeshie8

Answer 69

http://www.wolframalpha.com/input/?i=+%5B1%2F%282a%29%5Dln%5B%28a%2Bx%29%2F%28a-x%29%5D+%27

Answer 70

d/dx [ln(a+x) - ln(a-x) ] = 1/(a+x) + 1/(a-x) = 2a/(a^2 - x^2)

no multiply to 2a gives 1/(a^2-x^2)

Answer 71

Is "a" a constant or a variable?

Answer 72

constant

Answer 73

@sourwing but it's (1/2a)[1/(a+x) + 1/(a-x)]

Answer 74

no, 1/(2a) * [2a/(a^2-x^2)] = 1/(a^2 - x^2)

Answer 75

where did the 2a in the numerator come form i got -2x instead