# Derivative of a Natural Log

• Apr 20th 2008, 12:16 PM
Jeavus
Derivative of a Natural Log
I have the equation:

s(t) = ln((t^2 + 1)/(t^2 - 1)) + 6lnt

on the interval 1.1 < t < 10.

How do I find the derivative of this? If someone could show me step-by-step, I'd really appreciate it as I'm having troubles with this one.

Thanks!
• Apr 20th 2008, 12:26 PM
Moo
Hello,

Use the chain rule :

You have to find the derivative for ln((t^2 + 1)/(t^2 - 1))=ln(u(t))

Its derivative will be $\frac{u'(t)}{u(t)}$

So let's calculate u'(t).
This will be calculated with the quotient rule :

$u(t)=\frac{t^2+1}{t^2-1}=\frac{p(t)}{q(t)}$

$u'(t)=\frac{p'(t)q(t)-p(t)q'(t)}{(q(t))^2}$

Here, $p'(t)=2t$ and $q'(t)=2t$

Hence $u'(t)=\frac{2t(t^2-1)-2t(t^2+1)}{(t^2-1)^2}=\frac{2t^3-2t-2t^3-2t}{(t^2-1)^2}=\frac{-4t}{(t^2-1)^2}$

$s'(t)=\underbrace{\frac{1}{u(t)}}_{\frac{t^2-1}{t^2+1}} u'(t)+\frac{6}{t}$

$s'(t)=\frac{t^2-1}{t^2+1} \frac{-4t}{(t^2-1)^2}+\frac{6}{t}=\frac{-4t}{(t^2+1)(t^2-1)}+\frac{6}{t}$
• Apr 20th 2008, 12:35 PM
flyingsquirrel
Hi

$f(t) = \ln\frac{t^2+1}{t^2-1}+6\ln t=\ln (t^2+1)-\ln(t^2-1)+6\ln t$
and then differentiate each term:
$f'(t)=\frac{2t}{t^2+1}-\frac{2t}{t^2-1}+\frac{6}{t}=\frac{-4t}{(t^2+1)(t^2-1)}+\frac{6}{t}=\ldots$
• Apr 20th 2008, 12:36 PM
Moo
I was obsessed with the calculus... :D
• Apr 20th 2008, 12:40 PM
Jeavus
Thanks both of you.

Now, from that derivative, how do I find the maximum and minimum values for the function?

:)
• Apr 20th 2008, 12:44 PM
Jhevon
Quote:

Originally Posted by Jeavus
Thanks both of you.

Now, from that derivative, how do I find the maximum and minimum values for the function?

:)

set the derivative equal to zero and solve for t. you may need further tests (first or second derivative test) to classify the points as maximum or minimum
• Apr 20th 2008, 12:44 PM
Moo
Maximum and minimum values are points where the derivative is null.

Make the fractions in the derivative have the same denominator, then look the values of x for which the numerator is null.

To know if it's a maximum or a minimum, you can do a table of signs, or study the second derivative (which can be tricky)
• Apr 20th 2008, 12:47 PM
flyingsquirrel
Quote:

Originally Posted by Moo
I was obsessed with the calculus... :D

calculus or calculation ? :D
• Apr 20th 2008, 12:49 PM
Jeavus
Okay, I'm trying to solve for the t value but I can't.

s'(t) = (6t^4 - 4t^2 - 6)/(t)(t^2 + 1)(t^2 - 1)

Where do I go from here?
• Apr 20th 2008, 12:50 PM
Moo
Quote:

Originally Posted by flyingsquirrel

Avec le calcul direct de la dérivée

Is it better ? :D
I'm still not used to the difference between the two concepts... :(
• Apr 20th 2008, 12:51 PM
Jhevon
Quote:

Originally Posted by Jeavus
Okay, I'm trying to solve for the t value but I can't.

s'(t) = (6t^4 - 4t^2 - 6)/(t)(t^2 + 1)(t^2 - 1)

Where do I go from here?

ok, assuming that is the right answer, set the numerator equal to zero. note that it is quadratic in t^2
• Apr 20th 2008, 12:52 PM
Moo
Quote:

Originally Posted by Jeavus
Okay, I'm trying to solve for the t value but I can't.

s'(t) = (6t^4 - 4t^2 - 6)/(t)(t^2 + 1)(t^2 - 1)

Where do I go from here?

The derivative will be equal to 0 if the numerator is equal to 0 (and the denominator different of 0, or else it'll be undefined).

So substitute : u=t² :)