Check my answer for the derivative of $y=\tan^{-1}(40/h) -\tan^{-1}(32/h)$ 
Differentiate
$$y=\tan^{-1}{\frac {40}{h}}-\tan^{-1}{\frac {32}{h}}$$

My answer
Using the identity $\frac{d}{dy}\tan^{-1}(x)=\frac{1}{1+x^2}$, can I conclude that
$$\frac{dy}{dh}=\frac{1}{1+(\frac{40}{h})^2}(-\frac {40}{h^2})-\frac{1}{1+(\frac{32}{h})^2}(-\frac {32}{h^2})$$

It is correct, but you can simplify

$$
\begin{align*}
\frac{dy}{dh}&=\frac{-40}{(1+(\frac{40}{h})^2)h^2}-\frac{-32}{(1+(\frac{32}{h})^2)h^2}\\
&=\frac{-40}{40^2+h^2}+\frac{32}{32^2+h^2}\\
&=etc.
\end{align*}
$$


And you can also check "  WolframAlpha "
