Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

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

share|improve this question

4 Answers 4

up vote 1 down vote accepted

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 "

share|improve this answer

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

share|improve this answer
    
Thanks! a bunch. That was quick. I didn't simplify it was it was part of this massive question, just wanted to check on this particular step. –  Yellow Skies Oct 8 '12 at 11:51

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.

share|improve this answer

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

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.