<s>[INST] 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?
 [/INST] 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.)
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