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
$$\frac{dy}{dh}=\frac{1}{1+(\frac{40}{h})^2}(-\frac {40}{h^2})-\frac{1}{1+(\frac{32}{h})^2}(-\frac {32}{h^2})$$
 A: You are right. But the final result can be simplified a little bit.
A: 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 "
A: Well.. this sentence: "Using the identity $\frac{d}{dy}\tan^{-1}(x)=\frac{1}{1+x^2}$" is not correct in notation  as it stands.
We should look for $\frac{dy}{dh}$, and that is $\frac{dy}{dx}\cdot\frac{dx}{dh}$, exactly how you did it afterwards.
A: Why don't we use
$$\tan^{-1}\frac a h= \frac \pi 2-\cot^{-1}\frac a h=\frac \pi 2-\tan^{-1}\frac h a.$$
So, $$\frac{d(\tan^{-1}\frac a h)}{dh}=\frac{d(\frac \pi 2-\tan^{-1}\frac h a)}{dh}=-\frac{d(\tan^{-1}\frac h a)}{dh}=-\frac{1}{1+(\frac h a)^2}\frac 1 a=-\frac{a}{h^2+a^2}$$
So, $$\frac{d(\tan^{-1}\frac {40} h)}{dh}=-\frac{40}{h^2+40^2}$$ and 
$$\frac{d(\tan^{-1}\frac {32} h)}{dh}=-\frac{32}{h^2+32^2}$$
So, $$\frac{dy}{dh}=-\frac{40}{h^2+40^2}-\left(-\frac{32}{h^2+32^2}\right)=\frac{32}{h^2+32^2}-\frac{40}{h^2+40^2}=\frac{32\cdot 40\cdot 8-8h^2}{(h^2+32^2)(h^2+40^2)}$$
