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!
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!
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 $\displaystyle u'(t)=\frac{2t(t^2-1)-2t(t^2+1)}{(t^2-1)^2}=\frac{2t^3-2t-2t^3-2t}{(t^2-1)^2}=\frac{-4t}{(t^2-1)^2}$
$\displaystyle s'(t)=\underbrace{\frac{1}{u(t)}}_{\frac{t^2-1}{t^2+1}} u'(t)+\frac{6}{t}$
$\displaystyle s'(t)=\frac{t^2-1}{t^2+1} \frac{-4t}{(t^2-1)^2}+\frac{6}{t}=\frac{-4t}{(t^2+1)(t^2-1)}+\frac{6}{t}$
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:
$\displaystyle f'(t)=\frac{2t}{t^2+1}-\frac{2t}{t^2-1}+\frac{6}{t}=\frac{-4t}{(t^2+1)(t^2-1)}+\frac{6}{t}=\ldots$
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)
calculus or calculation ?