# can u fin the derivative of this function

• Jan 24th 2010, 03:07 AM
geriniki
can u fin the derivative of this function
i need the derivative of :
• Jan 24th 2010, 03:11 AM
tonio
Quote:

Originally Posted by geriniki
i need the derivative of :

First simplify the power: $\displaystyle \frac{1}{x}\left(x-\frac{1}{x}\right)=1-\frac{1}{x^2}$ , and now apply the chain rule to $\displaystyle \left(e^{f(x)}\right)'=f'(x)\cdot e^{f(x)}$ ...

Tonio
• Jan 24th 2010, 03:28 AM
drumist
The solution you provided doesn't match the problem.

Can I ask what solution you arrived at?
• Jan 24th 2010, 03:30 AM
geriniki
this must be solution it's in the textbook :/
sorry i haven't written it correctly = x - 1/x isn't in the power
• Jan 24th 2010, 03:34 AM
drumist
Is this the correct function?

$\displaystyle f(x) = e^{1/x}\left(x-\frac{1}{x}\right)$
• Jan 24th 2010, 03:39 AM
geriniki
yes
• Jan 24th 2010, 03:51 AM
drumist
You may use the product rule:

$\displaystyle f'(x) = e^{1/x} \cdot \left(x-\frac{1}{x}\right)' + \left(e^{1/x}\right)' \cdot \left(x-\frac{1}{x}\right)$

Can you find the derivatives of $\displaystyle x-\frac{1}{x}$ and $\displaystyle e^{1/x}$?
• Jan 24th 2010, 03:59 AM
geriniki
yes thank you
• Jan 24th 2010, 06:15 AM
HallsofIvy
Do you see why Tonio was confused? (Because he never gives a wrong answer!)

What you gave in your initial post was $\displaystyle e^{\frac{1}{x}\left(x- \frac{1}{x}\right)}$ not $\displaystyle e^{\frac{1}{x}}\left(x- \frac{1}{x}\right)$.