# Thread: Derivative of a Natural Log

1. ## 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!

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

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

$s'(t)=\frac{t^2-1}{t^2+1} \frac{-4t}{(t^2-1)^2}+\frac{6}{t}=\frac{-4t}{(t^2+1)(t^2-1)}+\frac{6}{t}$

3. 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:
$f'(t)=\frac{2t}{t^2+1}-\frac{2t}{t^2-1}+\frac{6}{t}=\frac{-4t}{(t^2+1)(t^2-1)}+\frac{6}{t}=\ldots$

4. I was obsessed with the calculus...

5. Thanks both of you.

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

6. 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

7. 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)

8. Originally Posted by Moo
I was obsessed with the calculus...
calculus or calculation ?

9. 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?

10. Originally Posted by flyingsquirrel
Avec le calcul direct de la dérivée

Is it better ?
I'm still not used to the difference between the two concepts...

11. 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

12. 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