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**

$\displaystyle f(x)=(\ln x+2x)^{1/3}$

Then, by the Generalized Exponent Rule,

$\displaystyle f'(x)=(1/3)(1/x+2)(\ln x+2x)^{-2/3}$

But $\displaystyle f'(x)=0$

Thus,

$\displaystyle (1/3)(1/x+2)(\ln x+2x)^{-2/3}=0$

thus,

$\displaystyle (1/x+2)=0$

because $\displaystyle (\ln x+2x)^{-2/3}$ is positive.

Thus,

$\displaystyle x=-1/2$

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 $\displaystyle (\ln x+2x)^{-2/3}$ 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:

$\displaystyle \frac{d}{dx}(x)e^y+xe^y\frac{dy}{dx}+\frac{d}{dx}( \ln x)y+\ln x\frac{dy}{dx}=\frac{d}{dx}(2)$

Simplify,

$\displaystyle e^y+xe^y\frac{dy}{dx}+\frac{1}{x}y+\ln x\frac{dy}{dx}=0$

Reorganize,

$\displaystyle xe^y\frac{dy}{dx}+\ln x\frac{dy}{dx}=-e^y-\frac{y}{x}$

Now factor,

$\displaystyle \frac{dy}{dx}(xe^y+\ln x)=-e^y-\frac{y}{x}$

Remember the problem says that point $\displaystyle (x,y)=(1,\ln 2)$

Substitute those values for $\displaystyle x,y$

Thus, (remember $\displaystyle e^{\ln 2}=2,\ln 1=0$)

$\displaystyle \frac{dy}{dx}((1)(2)+0)=-2-\ln 2$

Thus,

$\displaystyle \frac{dy}{dx}=\frac{-2-\ln 2}{2}$