1. ## Help Natural Logs

solve the equation f'(x) = 0

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

thanks

2. Originally Posted by thelazyman

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!

3. ## 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

4. 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>?

5. 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.

6. 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,
$\frac{dy}{dx}=\frac{-2-\ln 2}{2}$