Derivative of:
$$\left(x-a\right)\arctan\left(\dfrac{\left(y-b\right)\left(z-c\right)}{\left(x-a\right)\sqrt{\left(x-a\right)^2+\left(y-b\right)^2+\left(z-c\right)^2}}\right)$$
w.r.t. $c$ is, according to online calculator:
$$-\dfrac{\left(x-a\right)^2\left(y-b\right)}{\left(\left(c-z\right)^2+\left(x-a\right)^2\right)\sqrt{\left(x-a\right)^2+\left(y-b\right)^2+\left(z-c\right)^2}}$$
But I am getting: $$- \dfrac{1}{\sqrt{\left(x-a\right)^2+\left(y-b\right)^2+\left(z-c\right)^2}} \left[ \dfrac{\left(x-a\right)^2\left(y-b\right) \left[ \left( x-a \right)^2+\left(y-b\right)^2 \right]}{(x-a)^2 \left[ \left( x-a \right)^2+\left(y-b\right)^2+\left(c-z\right)^2 \right]+ (y-b)^2 (z-c)^2 } \right]$$
Why is this so? Are the both expressions equivalent?
Make substitutions to make this more manageable.
First, note that the derivative of the inverse tangent of a quotient simplifies a bit: $$ \biggl[ \arctan \frac{u}{v} \biggr]' = \frac{\bigl( \frac{u}{v} \bigr)'}{1 + \bigl( \frac{u}{v} \bigr)^2} = \frac{\frac{u'v - uv'}{v^2}}{1 + \frac{u^2}{v^2}} = \frac{u'v - uv'}{u^2 + v^2} $$
Now, set $X = x - a$, $Y = y - b$, and $Z = z - c$, and $R^2 = X^2 + Y^2 + Z^2$. Letting prime notation denote derivative with respect to $c$, $$ X' = Y' = 0, \quad Z' = -1, \quad\text{and}\quad R' = -\frac{Z}{R} $$ You want to find \begin{align} \biggl[ X \arctan \frac{YZ}{XR} \biggr]' &= X \, \frac{(YZ)'(XR) - (YZ)(XR)'}{(YZ)^2 + (XR)^2} \\& = X \, \frac{-YXR + YZX\frac{Z}{R}}{(YZ)^2 + (XR)^2} \\ &= -\frac{X^2 Y}{R} \, \frac{R^2 - Z^2}{Y^2Z^2 + X^2R^2} \\ &= -\frac{X^2 Y}{R} \, \frac{X^2 + Y^2}{Y^2Z^2 + X^2R^2}. \end{align} which agrees with your answer.
To recover the online calculator answer, expand $R^2$ and factor the denominator: $$ Y^2Z^2 + X^2R^2 = Y^2Z^2 + X^2(X^2 + Y^2 + Z^2) = (Z^2 + X^2)(X^2 + Y^2), $$ so the derivative looks like \begin{align} &= -\frac{X^2 Y}{R} \, \frac{X^2 + Y^2}{Y^2Z^2 + X^2R^2} \\ &= -\frac{X^2 Y}{R} \, \frac{X^2 + Y^2}{(Z^2 + X^2)(X^2 + Y^2)} \\ &= -\frac{X^2 Y}{(Z^2 + X^2)R}. \end{align}