can u fin the derivative of this function

**geriniki** · January 24th 2010, 03:07 AM

i need the derivative of :

**tonio** · January 24th 2010, 03:11 AM

Originally Posted by geriniki

i need the derivative of :

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

Tonio

**drumist** · January 24th 2010, 03:28 AM

The solution you provided doesn't match the problem.

Can I ask what solution you arrived at?

**geriniki** · January 24th 2010, 03:30 AM

this must be solution it's in the textbook :/
sorry i haven't written it correctly = x - 1/x isn't in the power

**drumist** · January 24th 2010, 03:34 AM

Is this the correct function?

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

**geriniki** · January 24th 2010, 03:39 AM

yes

**drumist** · January 24th 2010, 03:51 AM

You may use the product rule:

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

Can you find the derivatives of $x-\frac{1}{x}$ and $e^{1/x}$ ?

**geriniki** · January 24th 2010, 03:59 AM

yes thank you

**HallsofIvy** · January 24th 2010, 06:15 AM

Do you see why Tonio was confused? (Because he never gives a wrong answer!)

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

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