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.

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 $\dot\theta=-\frac7{75}$.
