Wrong answer in Thomas Calculus 14th Edition textbook There is this question on derivatives to which the answer is given as $\frac{43}{75}$rad/sec in the answers section. This answer appears to be wrong.

My Solution
$$
\theta+\tan^{-1}\left(\frac{6}{4-x}\right)+\tan^{-1}\left(\frac{3}{x}\right)=\pi
$$
which gets reduced to
$$
\theta=\pi-\tan^{-1}\left(\frac{3x+12}{4x-x^2-18}\right)
$$
Therefore,
$$
\frac{d\theta}{dt}=\left(\frac{3x^2+24x-102}{(4x-x^2+18)^2+(3x+12)^2}\right)\frac{dx}{dt}
$$
Given $x=4$ and $\frac{dx}{dt}=2\text{ cm/sec}$,
$$
\frac{d\theta}{dt}=-\frac{7}{75}\text{rad/sec}
$$
I've got the graph of $\theta$ as a function of $x$ here, which also indicates my answer is correct. Or am I? Kindly help.
 A: Here's an alternative explicit calculation.
Let $D$ be the end of the $6$-cm segment opposite $C.$
Let $E$ be the end of the $3$-cm segment opposite $A.$
The angle $\theta$ is the sum of the angles $\angle BDC$ and $\angle BEA.$
We can see that when $x=4,$ then $B$ has just reached $C$ and is moving perpendicular to the $6$ cm segment $BC$ at the rate $2$ cm per second,
so the angular velocity of $B$ around the point $D$ is
$2/6 = \frac13$ cm per second clockwise
and therefore $\angle BDC$ is changing at the rate
$-\frac13$ radian per second.
Also when $x = 4,$ the distance from $B$ to $E$ is $5$ cm.
Consider the radial and tangential components of the velocity vector of $B$
relative to the circle of radius $5$ around $E.$
The velocity vector and its components make a $3,4,5$ right triangle
with the magnitude of the tangential component equal to $\frac35$ the magnitude of the velocity; that is, the tangential component is
$\frac35\cdot 2 = \frac65$ cm per second clockwise.
This means that the angle $\angle BEA$ is increasing at the rate
$\frac{6/5}{5} = \frac{6}{25}$ cm per second.
So the answer is
$$ -\frac13 + \frac{6}{25} = - \frac{25}{75} + \frac{18}{25} = - \frac{7}{75}, $$
exactly as you found.
Note that
$$ \frac{25}{75} + \frac{18}{25} = \frac{43}{75}, $$
so the supposed answer in the book appears to be the result of a sign error.
A: The book's answer of $\frac{43}{75}$ rad/s obviously can't be true.
When $x=4$ point $B$ coincides with $C$, and then the angle above $\theta$ on the diagram (that is, between the upper dashed line and $BC$) changes by exactly $2/6 =\frac13$ rad/s. The angle below $\theta$ is changing in the opposite direction (and by no more than $\frac25$ rad/s, since the distance to the lower right corner is $5$), so the absolute value of the correct answer is at most $\frac 13$.
However $\frac{43}{75}$ is clearly larger than $\frac13=\frac{25}{75}$.
Since commenters have confirmed your result with a different calculation, it looks very likely that it is correct.
A: Differentiating your formula
$$\theta+\arctan\frac{6}{4-x}+\arctan\frac{3}{x}=\pi$$
with respect to time gives
$$\dot\theta+\frac1{1+\left(\dfrac6{4-x}\right)^2}\cdot\frac6{({4-x})^2}\dot x+\frac1{1+\left(\dfrac3x\right)^2}\cdot\frac{-3}{x^2}\dot x=0.$$
After substituting $\dot x=2$, this may be written
$$\dot\theta=6\left(\frac1{x^2+9}-\frac2{(4-x)^2+36}\right).$$
Now substituting $x=4$ yields the answer $\dot\theta=-\frac7{75}$.
