# Derivative of Natural Log

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• Mar 15th 2007, 01:17 PM
zachb
Derivative of Natural Log
Can someone please tell me what I'm doing wrong here, and show me how to do it correctly?

I need to differentiate g(x) = LN a-x/a+x

So,

g'(x) = 1/a-x/a+x *d/dx (a-x/a+x)
g'(x) = a+x/a-x *[(a+x)*d/dx(a-x) - (a-x)*d/dx(a+x)/(a+x)^2]
g'(x) = a+x/a-x *[(a+x)*(1-1) - (a-x)(1+1)/(a+x)^2]
g'(x) = a+x/a-x *[(a+x)*(0) - (a-x)(2)/(a+x)^2]
g'(x) = a+x/a-x *[0 - 2(a-x)/(a+x)^2]
g'(x) = a+x/a-x * -2a+2x/a^2+2ax+x^2
g'(x) = a+x*(-2a+2x)/a-x*(a^2+2ax+x^2)
g'(x) = -2a^2+2ax-2ax+2x^2/a^3+2a^2x+ax^2-a^2x-2ax^2-x^3)
g'(x) = -2a^2+2x^2/(a^3+a^2x-ax^2-x^3)

the answer is supposed to be g'(x) = -2a/a^2-x^2, but that's not what I get when I simplify . . .
• Mar 15th 2007, 01:29 PM
qbkr21
Re:
Here is the answer. I know it is right. The answer you have at the bottom is incorrect.

http://item.slide.com/r/1/38/i/L6mVO...x7ZkeiNuwW62H/
• Mar 15th 2007, 01:35 PM
zachb
My book says that the answer is g'(x) = -2a/a^2-x^2, but I'll look over your work.

The problem is g(x) = the natural log of a-x/a+x, not the natural log of a-x/ a+x, so I think you may ahve done it wrong . . .
• Mar 15th 2007, 01:48 PM
qbkr21
Quote:

Originally Posted by zachb
Can someone please tell me what I'm doing wrong here, and show me how to do it correctly?

I need to differentiate g(x) = LN a-x/a+x

So,

g'(x) = 1/a-x/a+x *d/dx (a-x/a+x)
g'(x) = a+x/a-x *[(a+x)*d/dx(a-x) - (a-x)*d/dx(a+x)/(a+x)^2]
g'(x) = a+x/a-x *[(a+x)*(1-1) - (a-x)(1+1)/(a+x)^2]
g'(x) = a+x/a-x *[(a+x)*(0) - (a-x)(2)/(a+x)^2]
g'(x) = a+x/a-x *[0 - 2(a-x)/(a+x)^2]
g'(x) = a+x/a-x * -2a+2x/a^2+2ax+x^2
g'(x) = a+x*(-2a+2x)/a-x*(a^2+2ax+x^2)
g'(x) = -2a^2+2ax-2ax+2x^2/a^3+2a^2x+ax^2-a^2x-2ax^2-x^3)
g'(x) = -2a^2+2x^2/(a^3+a^2x-ax^2-x^3)

the answer is supposed to be g'(x) = -2a/a^2-x^2, but that's not what I get when I simplify . . .

With your notation this is what you told me. I just went by what you said. Try using parenthesis. This is what you told me:

http://item.slide.com/r/1/102/i/kt1F...2GeQXISr0ogCN/
• Mar 15th 2007, 02:11 PM
frenzy
ln[(a-x)/(a+x)]=ln(a-x)-ln(a+x)

take the derivative

-1/(a-x)-1/(a+x)=-2a/(a^2-x^2)
• Mar 15th 2007, 02:27 PM
qbkr21
Quote:

Originally Posted by frenzy
ln[(a-x)/(a+x)]=ln(a-x)-ln(a+x)

take the derivative

-1/(a-x)-1/(a+x)=-2a/(a^2-x^2)

When you take the derivative of ln(a-x)-ln(a+x) you do not get 2a/(a^2-x^2)

you get:

http://item.slide.com/r/1/191/i/EZJp...x_b3nqpiz4mrp/

The way that first problem was posted was terrible absolutely terrible notation.
• Mar 15th 2007, 02:56 PM
frenzy
Quote:

Originally Posted by qbkr21
When you take the derivative of ln(a-x)-ln(a+x) you do not get 2a/(a^2-x^2)

you get:

http://item.slide.com/r/1/191/i/EZJp...x_b3nqpiz4mrp/

The way that first problem was posted was terrible absolutely terrible notation.

You might want to retry that derivative again.

What I wrote is correct.
• Mar 15th 2007, 03:38 PM
qbkr21
Re:
I am not going to argue with you but the derivative of ln(f(x))= (1/(f(x))* f'(x):cool:
• Mar 15th 2007, 03:41 PM
zachb
Frenzy, can you please show your work so that I can see how you got that answer? I already know what the answer should be, what I'm interested in is how you got there. :)
• Mar 15th 2007, 03:55 PM
frenzy
Quote:

Originally Posted by qbkr21
I am not going to argue with you but the derivative of ln(f(x))= (1/(f(x))* f'(x):cool:

This is correct...BUT, how you took the derivative is NOT correct.
Again...look at what you have written. It is wrong.

Quote:

Originally Posted by zachb
Frenzy, can you please show your work so that I can see how you got that answer? I already know what the answer should be, what I'm interested in is how you got there. :)

d/dx ln(a-x)= [1/(a-x)]*d/dx[(a-x)]=1/(a-x)]*(-1)=-1/(a-x)

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

-1/(a-x)-1/(a+x)=(-1(a+x)-1(a-x))/[(a-x)*(a+x)]

=(-a-x-a+x))/[a^2+ax-ax-x^2]=(-a-a)/(a^2-x^2)=-2a/(a^2-x^2)
• Mar 15th 2007, 04:04 PM
qbkr21
Re:
If you go to my second post where I show the problem aligned in the calculator. Then if you take the derivative of whats on the screen and put it on paper you get my exact answerr. Now maybe this whole scenario revolves around the fact that I didn't understand the way:

g(x)= ln x-a/x+a

There again this is terrible notation because it could me several different things:

g(x)= ln(x-a)/x+a ???

g(x)= ln(x)-a/x+a ???

or g(x)= ln(x-a/x+a) ???

I mean if my professor handed me this in class I would hand it right back to him simply because this in its entirety is "mathematically incorrect". A professor who does this obviously should be teaching grade school or not at all.
• Mar 15th 2007, 04:22 PM
frenzy
Quote:

Originally Posted by qbkr21
If you go to my second post where I show the problem aligned in the calculator. Then if you take the derivative of whats on the screen and put it on paper you get my exact answerr. Now maybe this whole scenario revolves around the fact that I didn't understand the way:

g(x)= ln x-a/x+a

There again this is terrible notation because it could me several different things:

g(x)= ln(x-a)/x+a ???

g(x)= ln(x)-a/x+a ???

or g(x)= ln(x-a/x+a) ???

I mean if my professor handed me this in class I would hand it right back to him simply because this in its entirety is "mathematically incorrect". A professor who does this obviously should be teaching grade school or not at all.

no doubt that the problem was missing some prentices

Going back to your 2nd post you have..

d/dx(a-x)=1-1=0
and
d/dx(a+x)=1+1=2
These are not correct.

you should have

d/dx(a-x)=0-1=-1
and
d/dx(a+x)=0+1=1

• Mar 15th 2007, 05:59 PM
zachb
Quote:

Originally Posted by frenzy
no doubt that the problem was missing some prentices

Going back to your 2nd post you have..

d/dx(a-x)=1-1=0
and
d/dx(a+x)=1+1=2
These are not correct.

you should have

d/dx(a-x)=0-1=-1
and
d/dx(a+x)=0+1=1

I don't understand why d/dx(a-x)=0-1=-1
and
d/dx(a+x)=0+1=1

Is that because "a" is a constant term?
• Mar 15th 2007, 06:05 PM
frenzy
Quote:

Originally Posted by zachb
I don't understand why d/dx(a-x)=0-1=-1
and
d/dx(a+x)=0+1=1

Is that because "a" is NOT a constant term?

a is a constant therefore

d/dx(a)=0

The derivative of any constant is zero.
• Mar 15th 2007, 06:12 PM
zachb
Okay, but I don't understand how we're supposed to know that "a" IS a constant term, and x is not. :( Is there some kind of rule for determining which term is constant when you have two or more varaibles? I thought that constant terms were just integers. For example, in f(x)=x^2-7x+5 the constant term is 5 and when you find f'(x) it's f'(x)=2x-7+0.:confused:
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