# Hard to slove

Who can slove this one:

f(x) = [ln(x) {{1-e^(3x)}^3}] / [{1+e^(3x)}^3x]

f '(x) = ¿¿¿¿??????? :surprised :surprised :surprised :surprised

## Answers and Replies

cristo
Staff Emeritus
Again, please show your work for this homework question. What methods of differentiation do you know of?

alba_ei,

This is not difficult but a little bit long to write.
What is your objective?
Is it for some homework, or do you have a practical application?

For an homework, the result would not be helpful for you, only the method matters.
There are only known functions in this expression: products, divisions, logarithm, exponential.
Reading a table of derivatives rules and a bit of patience is enough.
On wiki you can find the basis about derivatives and the http://en.wikipedia.org/wiki/Derivative" [Broken].

For a practical application, more details would be needed to decide how to proceed for the best result.
If this derivative is the only one in the project, then using a software like Mathematica could avoid any typing error.
If you only need numerical results, then "Numerical Recipes" explains what to care for.

Michel

Last edited by a moderator:
alba_ei,

As lalbatros suggests, this may be long and messy. But consider breaking this down into managable parts. For example, let your function be

f(x) = ln(x)*R(x)

where R(x) = P(x)/Q(x)

Then, apply the various rules of differentiation (product, quotient, etc) to perform the derivative. Start simply, and break each component down.

f'(x) = [ln(x)]' * R(x) + ln(x) *R'(x)

In the end, you should be able to find an expression for f'(x) in the form

f'(x) = R(x) * W(x)

where W(x) = 1/x + ln(x) * s(x) (you find s(x))

Give it a try, and come back if you still get stuck.

well, I saw that derivate on some past exam so the last night I remmember it and post it, just for curiosity because I can't slove it.

f(x) = ln(x)*R(x)

where R(x) = P(x)/Q(x)

I get stuck when I try to get the derivate of R(x).

cristo
Staff Emeritus
Use the quotient rule: $$R'(x)=\frac{Q(x)P'(x)-P(x)Q'(x)}{Q(x)^2}$$