Derivative of ln
- Thread starter jtcook95
- Start date
Show us what you have done, or what you have tried . . .
There is no question indicated (except the Title line says derivative), nor any indication of whether you are using the limit definition, or the chain rule and the quotient rule.
HallsofIvy
Elite Member
- Joined
- Jan 27, 2012
- Messages
- 7,763
73!
D
Deleted member 4993
Guest
Incorrect.... it is 42..... it is always 42 !!!
HallsofIvy
Elite Member
- Joined
- Jan 27, 2012
- Messages
- 7,763
Now, those are fighting words!
I'll give it a shot:
\(\displaystyle y = \dfrac{1}{2} \ln (\dfrac{1 + x}{1 - x})\)
\(\displaystyle y' = \dfrac{1}{2} [(\dfrac{1}{u}) du]\)
\(\displaystyle y' = \dfrac{1}{2} [(\dfrac{1}{u}) \dfrac{(1 - x)(1) - (1 + x)(-1)}{(1 - x)^{2}}]\)
\(\displaystyle y' = \dfrac{1}{2} [(\dfrac{1}{u}) \dfrac{(1 - x) - (1 - x)}{(1 - x)^{2}}]\)
\(\displaystyle y' = \dfrac{1}{2} [(\dfrac{1}{u}) \dfrac{(1 - x) - (1 - x)}{(1 - x)(1 - x)}]\)
\(\displaystyle y' = \dfrac{1}{2} [(\dfrac{1}{u}) \dfrac{1}{1 - x} - \dfrac{1}{1 - x}]\)
\(\displaystyle y' = \dfrac{1}{2} [(\dfrac {1}{\dfrac{1 + x}{1 - x}}) \dfrac{1}{1 - x} - \dfrac{1}{1 - x}]\)
\(\displaystyle y = \dfrac{1}{2} \ln (\dfrac{1 + x}{1 - x})\)
\(\displaystyle y' = \dfrac{1}{2} [(\dfrac{1}{u}) du]\)
\(\displaystyle y' = \dfrac{1}{2} [(\dfrac{1}{u}) \dfrac{(1 - x)(1) - (1 + x)(-1)}{(1 - x)^{2}}]\)
\(\displaystyle y' = \dfrac{1}{2} [(\dfrac{1}{u}) \dfrac{(1 - x) - (1 - x)}{(1 - x)^{2}}]\)
\(\displaystyle y' = \dfrac{1}{2} [(\dfrac{1}{u}) \dfrac{(1 - x) - (1 - x)}{(1 - x)(1 - x)}]\)
\(\displaystyle y' = \dfrac{1}{2} [(\dfrac{1}{u}) \dfrac{1}{1 - x} - \dfrac{1}{1 - x}]\)
\(\displaystyle y' = \dfrac{1}{2} [(\dfrac {1}{\dfrac{1 + x}{1 - x}}) \dfrac{1}{1 - x} - \dfrac{1}{1 - x}]\)
Last edited:
In response to lookagain,
The problem calls for the quotient rule \(\displaystyle \dfrac{g(f') - f(g')}{g^{2}}\) given \(\displaystyle \dfrac{f}{g}\) NOT the product rule \(\displaystyle g(f') + f(g')\) given \(\displaystyle f(g)\) so why the plus sign in there?
\(\displaystyle y = \dfrac{1}{2} \ln (\dfrac{1 + x}{1 - x})\)
\(\displaystyle y' = \dfrac{1}{2} [(\dfrac{1}{u}) du]\)
\(\displaystyle y' = \dfrac{1}{2} [(\dfrac{1}{u}) \dfrac{(1 - x)(1) - (1 + x)(-1)}{(1 - x)^{2}}]\)
\(\displaystyle y' = \dfrac{1}{2} [(\dfrac{1}{u}) \dfrac{(1 - x) - (1 - x)}{(1 - x)^{2}}]\)
\(\displaystyle y' = \dfrac{1}{2} [(\dfrac{1}{u}) \dfrac{1 - x - 1 + x}{(1 - x)^{2})}]\)
We a change to a plus sign here. But not two changes to a plus sign as you wrote.
\(\displaystyle y' = \dfrac{1}{2} [(\dfrac{1}{u}) \dfrac{0}{(1 - x)^{2})}]\)
\(\displaystyle y' = \dfrac{1}{2} [(\dfrac{1}{u}) (0)]\)
\(\displaystyle y' = \dfrac{1}{2} [(\dfrac{1}{\dfrac{1 + x}{1 - x}}) (0)]\)
\(\displaystyle y' = 0\)
The problem calls for the quotient rule \(\displaystyle \dfrac{g(f') - f(g')}{g^{2}}\) given \(\displaystyle \dfrac{f}{g}\) NOT the product rule \(\displaystyle g(f') + f(g')\) given \(\displaystyle f(g)\) so why the plus sign in there?
\(\displaystyle y = \dfrac{1}{2} \ln (\dfrac{1 + x}{1 - x})\)
\(\displaystyle y' = \dfrac{1}{2} [(\dfrac{1}{u}) du]\)
\(\displaystyle y' = \dfrac{1}{2} [(\dfrac{1}{u}) \dfrac{(1 - x)(1) - (1 + x)(-1)}{(1 - x)^{2}}]\)
\(\displaystyle y' = \dfrac{1}{2} [(\dfrac{1}{u}) \dfrac{(1 - x) - (1 - x)}{(1 - x)^{2}}]\)
\(\displaystyle y' = \dfrac{1}{2} [(\dfrac{1}{u}) \dfrac{1 - x - 1 + x}{(1 - x)^{2})}]\)
We a change to a plus sign here. But not two changes to a plus sign as you wrote.\(\displaystyle y' = \dfrac{1}{2} [(\dfrac{1}{u}) \dfrac{0}{(1 - x)^{2})}]\)
\(\displaystyle y' = \dfrac{1}{2} [(\dfrac{1}{u}) (0)]\)
\(\displaystyle y' = \dfrac{1}{2} [(\dfrac{1}{\dfrac{1 + x}{1 - x}}) (0)]\)
\(\displaystyle y' = 0\)
Last edited:
In response to lookagain,
The problem calls for the quotient rule \(\displaystyle \dfrac{g(f') - f(g')}{g^{2}}\) given \(\displaystyle \dfrac{f}{g}\) NOT the product rule \(\displaystyle g(f') + f(g')\) given \(\displaystyle f(g)\) so why the plus sign in there?
\(\displaystyle y = \dfrac{1}{2} \ln (\dfrac{1 + x}{1 - x})\)
\(\displaystyle y' = \dfrac{1}{2} [(\dfrac{1}{u}) du]\)
\(\displaystyle y' = \dfrac{1}{2} [(\dfrac{1}{u}) \dfrac{(1 - x)(1) - (1 + x)(-1)}{(1 - x)^{2}}] \ \ \ \) <-------------
\(\displaystyle y' = \dfrac{1}{2} [(\dfrac{1}{u}) \dfrac{(1 - x) - (1 - x)}{(1 - x)^{2}}]\)
\(\displaystyle y' = \dfrac{1}{2} [(\dfrac{1}{u}) \dfrac{1 - x - 1 + x}{(1 - x)^{2})}]\)
\(\displaystyle y' = \dfrac{1}{2} [(\dfrac{1}{u}) \dfrac{0}{(1 - x)^{2})}]\)
\(\displaystyle y' = \dfrac{1}{2} [(\dfrac{1}{u}) (0)]\)
\(\displaystyle y' = \dfrac{1}{2} [(\dfrac{1}{\dfrac{1 + x}{1 - x}}) (0)]\)
\(\displaystyle y' = 0\)
Jason76,
look again at numerator portion of your step above at the arrow.
And contrast the correct algebra I'll show with your errors there:
\(\displaystyle (1 - x)(1) \ - \ (1 + x)(-1) \ = \)
\(\displaystyle 1 - x \ - \ (-1 - x) \ = \ \)
\(\displaystyle 1 - x \ + \ 1 + x \ = \)
2
--------------
You see? It does not equal 0.
HallsofIvy
Elite Member
- Joined
- Jan 27, 2012
- Messages
- 7,763
In the original post it was not clear whether the function was \(\displaystyle \frac{1}{2}\frac{ln(1+ x)}{1- x}\) or \(\displaystyle \frac{1}{2}ln\left(\frac{1+ x}{1- x}\right)\). Now, Jason76 seems to think it is the latter.
The first thing I would do is write it as \(\displaystyle y= \frac{1}{2}\left(ln(1+ x)- ln(1- x)\right)\). The derivative of that should be easy.
The first thing I would do is write it as \(\displaystyle y= \frac{1}{2}\left(ln(1+ x)- ln(1- x)\right)\). The derivative of that should be easy.