Help Natural Logs

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January 15th 2006, 02:51 PM
thelazyman

Help Natural Logs

Hi, please help

solve the equation f'(x) = 0

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

thanks
January 15th 2006, 03:32 PM
ThePerfectHacker

Quote:

Originally Posted by thelazyman

Hi, please help

solve the equation f'(x) = 0

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

thanks

Given the function,
$f(x)=(\ln x+2x)^{1/3}$
Then, by the Generalized Exponent Rule,
$f'(x)=(1/3)(1/x+2)(\ln x+2x)^{-2/3}$
But $f'(x)=0$
Thus,
$(1/3)(1/x+2)(\ln x+2x)^{-2/3}=0$
thus,
$(1/x+2)=0$
because $(\ln x+2x)^{-2/3}$ is positive.
Thus,
$x=-1/2$

This is my 2:):) Post!
January 15th 2006, 04:12 PM
thelazyman

Natural 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
January 15th 2006, 04:14 PM
thelazyman

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>?
January 15th 2006, 05:05 PM
ThePerfectHacker

Quote:

Originally Posted by thelazyman

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>?

You could think of it like that.
See the problem is that $(\ln x+2x)^{-2/3}$ is never zero thus you can divide both sides by this.
January 15th 2006, 05:26 PM
ThePerfectHacker

Quote:

Originally Posted by thelazyman

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

Take the derivative of both sides implicitly,
Remember to use the product rule :eek:
$\frac{d}{dx}(x)e^y+xe^y\frac{dy}{dx}+\frac{d}{dx}( \ln x)y+\ln x\frac{dy}{dx}=\frac{d}{dx}(2)$
Simplify,
$e^y+xe^y\frac{dy}{dx}+\frac{1}{x}y+\ln x\frac{dy}{dx}=0$
Reorganize,
$xe^y\frac{dy}{dx}+\ln x\frac{dy}{dx}=-e^y-\frac{y}{x}$
Now factor,
$\frac{dy}{dx}(xe^y+\ln x)=-e^y-\frac{y}{x}$
Remember the problem says that point $(x,y)=(1,\ln 2)$
Substitute those values for $x,y$
Thus, (remember $e^{\ln 2}=2,\ln 1=0$ )
$\frac{dy}{dx}((1)(2)+0)=-2-\ln 2$
Thus,
$\frac{dy}{dx}=\frac{-2-\ln 2}{2}$