Help Natural Logs

**thelazyman** · January 15th 2006, 02:51 PM

Hi, please help

solve the equation f'(x) = 0

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

thanks

**ThePerfectHacker** · January 15th 2006, 03:32 PM

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!

**thelazyman** · January 15th 2006, 04:12 PM

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

**thelazyman** · January 15th 2006, 04:14 PM

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>?

**ThePerfectHacker** · January 15th 2006, 05:05 PM

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.

**ThePerfectHacker** · January 15th 2006, 05:26 PM

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

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

Thread: Help Natural Logs

LinkBack

Thread Tools

Display

Help Natural Logs

Natural Logs again

Similar Math Help Forum Discussions

fun with natural logs....

Few HW problems involving Logs, Natural logs, and Continuity.

Natural logs????

Dealing with Logs and Natural Logs

several questions-logs/natural logs

Search Tags