The point on $y = \sqrt{\ln x}$ closest to $(4,0)$ Find the point of the graph of $y = \sqrt{\ln x}$ that is closest to the point $(4,0)$.
I started by using the distance formula and plugging in the values where:
$$d = \sqrt{(x-4)^2 + (\sqrt{\ln x} - 0)^2}$$
$$d = \sqrt{(x^2-8x+16+\ln x)}$$
I know we should set this equal to 0.
Stuck here do I somehow combine ln x with 8x?
 A: After writing this out using the distance formula, which you have done, you need to take the derivative. Then, you can set the derivative equal to zero and solve for the critical values to optimize the original equation. 
A: You don't want to set the distance itself to $0$, but rather the rate of change of the distance (as a function of $x$). So using the formula you correctly derived: $$\begin{array}{rcl} d(x) & = & \sqrt{(x-4)^2+\ln x} \\ & = & \left[(x-4)^2 + \ln x \right]^{\frac{1}{2}} \end{array} $$ we now differentiate it (using the chain rule several times) to get: $$\begin{array}{rcl} d'(x) & = & \frac{1}{2}\left[(x-4)^2 + \ln x \right]^{-\frac{1}{2}} \cdot \left[ 2(x-4) + \frac{1}{x} \right] .  \end{array}$$ Now setting this equal to zero, we find that (since $ab=0$ implies that $a=0$ or $b=0$, and conversely) our problem can be reduced to finding the $x$ which satisfy the following condition: $$\frac{1}{2\sqrt{(x-4)^2 + \ln x }}=0 \quad \text{or} \quad 2(x-4)+\frac{1}{x}=0. $$ Since there are no $x$ satisfying the left equation (since there is no number $t$ such that $\frac{1}{t}=0$), we just have to find the $x$ such that : $$2(x-4)+\frac{1}{x}=0 \implies 2x^2 -8x + 1 =0.  $$ Now you can just use the quadratic formula to find two candidate values of $x$, call them $x_1$ and $x_2$. Then test to see which of the two points $(x_1, \sqrt{\ln x_1})$ or $(x_2, \sqrt{\ln x_2})$ works. (We might have introduced an extra solution when multiplying by $x$, so that's why we have to check both to see whether they are correct or not.)
