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I've seen this in Stewart calculus book:

$$\frac{\mathrm d \tan^{-1} x}{\mathrm dx} = \frac1{1+x^2}$$

But how do I get it? If I do it myself,

$$\frac{\mathrm d \tan^{-1} x}{\mathrm dx} = \frac{\mathrm d \frac{\cos x}{\sin x}}{\mathrm dx} = \frac{-1}{\sin^2 x}$$

How can he get rid of the trigonometric functions ($\sin$, $\cos$)?

Thanks!

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Thanks for editing and formatting it! –  user1531 Sep 6 '11 at 13:01
    
See this. –  J. M. Sep 6 '11 at 13:02
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The notation $\tan ^{-1}x$ refers to the compositional inverse of tangent (i.e. arctan), not the reciprocal! –  The Chaz 2.0 Sep 6 '11 at 13:07
    
Thank you guys, i was really wrong! –  user1531 Sep 6 '11 at 13:15
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2 Answers

up vote 8 down vote accepted

Arctan is the inverse operation of tan, in the sense that $\arctan \tan \theta = \theta$.

To prove this, we note that $y = \arctan x$ implies that $\tan y = x$. So that $$\rm \frac{d}{dx} \arctan x = \dfrac{1}{\frac{d \tan y}{dy}} = \frac{1}{\sec ^2 y} = \frac{1}{1 + \tan ^2 y} = \frac{1}{1 + x^2}$$

And that is exactly what your book says.


---EDIT---

Listening to Hardy's suggestion, I add the following. Suppose we have two functions, $f$ and $g$ s.t. $f(g(x)) = g(f(x)) = x$, i.e. they are inverses of each other. Then by the chain rule, differentiating $f(g(x)) = x$, we get $\rm\frac{df}{dg} \frac{dg}{dx} = 1$, and this implies that $\rm \frac{dg}{dx} = \dfrac{1}{\frac{df}{dg}}$. I used this implicitly in my transition from the first to the second parts of the equation above.

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The first step in the argument above is essentially the chain rule, and might bear more explanation. –  Michael Hardy Sep 6 '11 at 13:04
    
I need a few minutes to get it –  user1531 Sep 6 '11 at 13:06
    
@Michael: Thanks, sure thing. –  mixedmath Sep 6 '11 at 13:17
    
Personally I would have said if $y = \arctan x$ then $x= \tan y$ so $\frac{dx}{dy} = 1+\tan^2 y = 1+x^2$ leading to the result, but it is much the same thing. –  Henry Sep 6 '11 at 13:17
    
Thank you, i feel stupid. I confused cotangent with arctangent. –  user1531 Sep 6 '11 at 13:17
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Here $\tan^{-1}$ refers to $\arctan$, not to the multiplicative inverse.

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Yes, but arctan x = cos x / sen x am i wrong? –  user1531 Sep 6 '11 at 13:00
    
No (in the sense equality is not true), $\arctan$ has the following property $\tan(\arctan(x))=x$ for every $x\in\mathbb{R}$. So, that equality is not true. –  Josué Tonelli-Cueto Sep 6 '11 at 13:01
    
So? If you do it then you get the result i gave before: -1 / sen^2 x Not 1 / 1 + X² –  user1531 Sep 6 '11 at 13:02
    
We have said that $\arctan\neq\frac{\cos}{\sin}$; so no. –  Josué Tonelli-Cueto Sep 6 '11 at 13:04
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I don't know how many times a student has asked me: "Is arctangent the same as cotangent?". It's a pretty standard question. One who knows trigonometry know that the answer is no. –  Michael Hardy Sep 6 '11 at 13:09
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