Hi, please help
solve the equation f'(x) = 0
f(x)= (ln x +2x) ^ 1/3
thanks
Given the function,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!
Take the derivative of both sides implicitly,Originally Posted by thelazyman
Remember to use the product rule
$\displaystyle \frac{d}{dx}(x)e^y+xe^y\frac{dy}{dx}+\frac{d}{dx}( \ln x)y+\ln x\frac{dy}{dx}=\frac{d}{dx}(2)$
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,
$\displaystyle \frac{dy}{dx}=\frac{-2-\ln 2}{2}$