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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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    Thank you guys, i was really wrong!2011-09-06

2 Answers 2

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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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    It would be better in the first line to write $\tan(\arctan\theta)=\theta$. The current version is not quite right, take $\theta=2\pi$.2011-09-06
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Here $\tan^{-1}$ refers to $\arctan$, not to the multiplicative inverse.

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    If $y = \tan x$ then $x = \arctan y$, at least if $x$ is between $-\pi/2$ and $\pi/2$. That's how the arctangent is defined. The cotangent, on the other hand is cosine over sine.2011-09-06