Question 899462
Assuming the problem is {{{sqrt(t^2 + 2ln(t))}}}, we know that we are going to have to do the chain rule. I find it easiest to use simple substitution to do this. You'll see what I mean in a second.

Let's let u = t^2 + 2ln(t) so du = 2t + 2/t dt  [but dt is just 1 since derivative of t = 1]

Let's let v = sqrt(u) so dv = 1/2sqrt(u) * du

Going from dv we have

{{{dv = 1/2sqrt(u) * du}}}

Substitute du = 2t + 2/t

{{{1/(2*sqrt(u)) * (2t + 2/t)}}}

Substituting u in terms of t, we get 

{{{1/(2*sqrt(t^2+2ln(t))) * (2t + 2/t)}}}

Getting rid of 2.

{{{(t+1/t)/(sqrt(t^2+2ln(t)))}}}  <--- perfect acceptable

Getting the Wolfram-Alpha answer:

Get a common denominator in the numerator.

{{{((t^2+1)/t)/(sqrt(t^2+2ln(t)))}}}

Drop the t into the denominator

{{{(t^2+1)/(t*sqrt(t^2+2ln(t)))}}}

So, to recap, any time you have functions within functions, we just assign them variables.

Like  sin(cos(x)))

We let u = cos(x) so du = -sin(x)

Now we let v = sin(u) so dv = cos(u) du

We go from dv and get cos(u) * (-sin(x))

Substitute u back in and get cos(cos(x)) * -sin(x).

Clean it up a bit and get -cos(cos(x))*sin(x).

Hope this makes sense and makes the chain rule a little easier! Feel free to email me if it does not.  <swincher4391@yahoo.com>