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
$$\frac{dy}{dh}=\frac{1}{1+(\frac{40}{h})^2}(-\frac {40}{h^2})-\frac{1}{1+(\frac{32}{h})^2}(-\frac {32}{h^2})$$

You are right. But the final result can be simplified a little bit.
