Hi, please help

solve the equation f'(x) = 0

f(x)= (ln x +2x) ^ 1/3

thanks

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- Jan 15th 2006, 02:51 PMthelazymanHelp Natural Logs
Hi, please help

solve the equation f'(x) = 0

f(x)= (ln x +2x) ^ 1/3

thanks - Jan 15th 2006, 03:32 PMThePerfectHackerQuote:

Originally Posted by**thelazyman**

Then, by the Generalized Exponent Rule,

But

Thus,

thus,

because is positive.

Thus,

This is my 2:):) Post! - Jan 15th 2006, 04:12 PMthelazymanNatural Logs again
Hi, test is tommorow need help with this question

For each of the following, find the slope of the tangent.

For the curve defined by xe^y + ylnx = 2 at (1, ln2).

Thanks - Jan 15th 2006, 04:14 PMthelazyman
Hi, I understand how you got the -1/2, but what happens to the 1/3 and (lnx + 2x)^ -2/3, do they cancel out and why>?

- Jan 15th 2006, 05:05 PMThePerfectHackerQuote:

Originally Posted by**thelazyman**

See the problem is that is never zero thus you can divide both sides by this. - Jan 15th 2006, 05:26 PMThePerfectHackerQuote:

Originally Posted by**thelazyman**

*implicitly*,

Remember to use the product rule :eek:

Simplify,

Reorganize,

Now factor,

Remember the problem says that point

Substitute those values for

Thus, (remember )

Thus,