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!

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- Apr 20th 2008, 12:16 PMJeavusDerivative of a Natural LogI 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 PMMoo
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 $\displaystyle \frac{u'(t)}{u(t)}$

So let's calculate u'(t).

This will be calculated with the quotient rule :

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

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

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

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

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

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

Otherwise, you can also split the log into two parts :

$\displaystyle f(t) = \ln\frac{t^2+1}{t^2-1}+6\ln t=\ln (t^2+1)-\ln(t^2-1)+6\ln t$

and then differentiate each term:

$\displaystyle f'(t)=\frac{2t}{t^2+1}-\frac{2t}{t^2-1}+\frac{6}{t}=\frac{-4t}{(t^2+1)(t^2-1)}+\frac{6}{t}=\ldots$ - Apr 20th 2008, 12:36 PMMoo
I was obsessed with the calculus... :D

- Apr 20th 2008, 12:40 PMJeavus
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 PMJhevon
- Apr 20th 2008, 12:44 PMMoo
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 PMflyingsquirrel
calculus or calculation ? :D

- Apr 20th 2008, 12:49 PMJeavusOkay, 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 PMMoo
- Apr 20th 2008, 12:51 PMJhevon
- Apr 20th 2008, 12:52 PMMoo