# Thread: Differentiation Problem for a forum newbie :)

1. ## Differentiation Problem for a forum newbie :)

Hi,

I was wondering if someone could explain how I get from A to B on this problem? I know it involves the product rule, but I'm not sure how to correctly apply it?

I'm not sure how you display formulas, so I have attached a file.

2. General rule:

f(x)g(x) = f'(x)g(x) + f(x)g'(x)

So if you have tow functions, first take derivative of the first multiplied by the second function then add the function of the first times the derivative of the second.

Just like in your file is done.

3. Hi,

thanks for you reply - I'm still not totally clued up. Could you explain how this problem is broken down? I'm not sure why DS/DT ends up in the solution. What are the f'(x) g(x) f(x) g'(x) in the problem?

If you could explain that, I would really appreciate it :-)

4. Hi,

The total function where you want the derivative from is:
$\displaystyle pS_{T}e^{-iT}$

There are two functions of T in this expression:
$\displaystyle f(T) = S_{T}$
and
$\displaystyle g(T) = e^{-iT}$

Now use the rule and you should get the same result as in your document.

And dS/dT is just the derivative of $\displaystyle S_{T}$ with respect to T.

Let me know if you can take it from here!

Bye

5. I found your picture rather unsettling, but assumed it must be explained by some esoteric notation. However, if

Originally Posted by jeneverboy
$\displaystyle f(T) = S_{T}$
is right, then I think jeneverboy maybe just scanned the beginning and end, which are OK if we're to understand the derivative is set to zero, but didn't notice those unwanted (?) bracketed expressions in lines 2 and 3 - which are what I take it are confusing me and you, Karty. Typo? Where's it from?

6. thank you both for your help.

the problem is related to forrestry economics.

i think the brackets were due to a previous simplification if I understand correctly.

I've updated by document to show all calcs.

I think I understand now.

to confirm

F'(T)=DS/Dt and G'(T)=D/Dte^-iT

Thanks so much for the help.

Katy