http://img193.imageshack.us/img193/4...00122at428.png

I blanked on how to solve this. I thought i take the seventh root and put it inside, so it would be (x)^7/2 then differentiate ...but i was wrong. Any takers? Thanks!

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- Jan 22nd 2010, 12:39 PMEvan.Kimiadifferentiate this equation...
http://img193.imageshack.us/img193/4...00122at428.png

I blanked on how to solve this. I thought i take the seventh root and put it inside, so it would be (x)^7/2 then differentiate ...but i was wrong. Any takers? Thanks! - Jan 22nd 2010, 12:40 PMLord Darkin
Wouldn't it just be (1/(seventh root of x)) * Chain rule?

- Jan 22nd 2010, 01:31 PMdrumist
$\displaystyle \sqrt[7]{x} = x^{1/7}$

You can use chain rule. - Jan 22nd 2010, 01:37 PMdrumist
Or even easier:

$\displaystyle f(x) = \ln(\sqrt[7]{x}) = \ln(x^{1/7}) = \frac{1}{7} \ln(x)$

- Jan 22nd 2010, 02:16 PMEvan.Kimia
It was that part i forgot. thanks.

- Jan 23rd 2010, 03:51 AMSeulementrienanswer
f(x)=(1/7)lnx

so f'(x)=1/7x