Math Help - Derivatives...

1. Derivatives...

Ugh I understand all my basic differentiation rules and my basic algebra but I get into a problems when I have longer problems like these. Sorry if it seems like I'm asking you to do the problem for me but would anyone mind giving me a little guidance?

Differentiate and solve for T when dP/dT = 0 (A is a constant)

$
P=2(\frac{A-\frac{1}{3}T^2}{\frac{2}{3}T})+\frac{2T}{3}+2\sqrt {(\frac{T}{3})^{2}+T^{2}}
$

I just don't really know where to start or anything.. Thanks!

2. the trick here is to use some algebra to simplify your expression. before doing any calculus

$P=2 \left(\frac{A-\frac{1}{3}T^2}{\frac{2}{3}T} \right)+\frac{2T}{3}+2\sqrt{\left( \frac{T}{3}\right )^{2}+T^{2}}$

$\frac{4A}{3T} - T + \frac{2T}{3} + 2 \sqrt{T^2(\frac{1}{9} +1)}$

$\frac{4A}{3T} - \frac{T}{3} + 2T\frac{\sqrt{10}}{3}$

$\frac{4A}{3T} + T\frac{2\sqrt{10}-1}{3}$

Now you differentiate.

I am unsure about what "solve for T" means though.

Edit: I noticed you defined what "solve for T" means, I am sure you can finish the problem from here, post if you have any problems

3. Wow the power of simplification....... =p

Got a little further and then I have problems again solving for T.. here's what i have so far....

$
\frac{4A}{3T} + T\frac{2\sqrt{10}-1}{3}
$

dP/dT = (3A)(-1)(3T)^(-2)(3) + $\frac{2\sqrt{10}-1}{3}$

0 = (3A)(-1)(3T)^(-2)(3) + $\frac{2\sqrt{10}-1}{3}$

$-\frac{2\sqrt{10}-1}{3}$= (3A)(-3)(3T)^(-2)

Not really sure how to take it any further or if my previous differentiation is even correct.... Would appreciate any help. Thanks!

4. $P= \frac{4A}{3T} + T\frac{2\sqrt{10}-1}{3}
$

$\frac{dP}{dT} = \frac{-4A}{3T^2} + \frac{2\sqrt{10}-1}{3}$

if $\frac{dP}{dT} = 0$ we get

$\frac{4A}{3T^2} = \frac{2\sqrt{10}-1}{3}$

reciprocate and cancel the 3

$\frac{T^2}{4A} = \frac{1}{2\sqrt{10}-1}$

$T = \pm \sqrt{\frac{4A}{2\sqrt{10}-1}}$