# Check my answer: Differentiate $y=\tan^{-1}{\frac {40}{h}}-\tan^{-1}{\frac {32}{h}}$

Differentiate

$$y=\tan^{-1}{\frac {40}{h}}-\tan^{-1}{\frac {32}{h}}$$

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

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

?

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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 "

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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, $$\frac{dy}{dh}=-\frac{40}{h^2+40^2}-\left(-\frac{32}{h^2+32^2}\right)=\frac{32}{h^2+32^2}-\frac{40}{h^2+40^2}=\frac{32\cdot 40\cdot 8-8h^2}{(h^2+32^2)(h^2+40^2)}$$

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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.