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Prove $(\tan^{-1}{x})' = \frac{1}{1+x^2}$

How should I proceed? I tried:

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Then another which I think is wrong:

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What should I be doing?

UPDATE

In my lecture notes ... maybe a typo:

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  • 0
    Surely a typo in your lecture notes.2012-02-22

4 Answers 4

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I guess that you want to prove that (\tan^{-1} x)'=1/(1+x^2).

(\tan^{-1} x)'=\frac{1}{\tan' y}

where $\tan y=x$. Therefore

(\tan^{-1} x)'=\frac{1}{1+\tan^2 y}=\frac{1}{1+x^2}

5

The answer above is also correct, but for clarity here it goes

Let $y=\tan^{-1} x \Rightarrow \tan(y) = x$

Differentiate both sides:

Left side will be $\sec^2(y) \dfrac{dy}{dx}$, Right side will be $1$

But using the trigonometric identity $\sec^{2}(y) = 1 + \tan^{2}(y)$

$\begin{align} (1+\tan^{2}(y))&\frac{dy}{dx} = 1 \\ \Rightarrow& \frac{dy}{dx} = \frac{1}{1+\tan^{2}(y)} \end{align}$

Since we began with $y=\tan^{-1} x \Rightarrow \tan(y) = x$

$\frac{dy}{dx} = \frac{1}{1+x^2}$

But \dfrac{dy}{dx} = (\tan^{-1} x)', Therefore (\tan^{-1}x)' = \frac{1}{1+x^2}

  • 0
    A minor quibble: you shouldn't use expressions such as "the answer above..." because what was above at the time of your writing may not be above at the time of our reading :)2012-02-23
3

It should be (\tan^{-1}x)' (or arc$\tan x$). Then use $y=\tan^{-1} x$

$\Rightarrow \tan y= x$ Now differentiate both sides w.r.t. $x$ to get sec^2 y. y'=1, or (1+\tan^2 y).y'=1, or (1+x^2).y'=1

3

Has your class covered integration? It may be easier to prove that $\int\frac{dx}{1+x^2}=\tan^{-1}x+C$. To clear up the denominator, try the substitution

$x=\tan\theta,dx=\sec^2\theta d\theta$

$\int\frac{dx}{1+x^2}=\int\frac{\sec^2\theta d\theta}{1+\tan^2\theta}=\int\frac{\sec^2\theta d\theta}{\sec^2\theta}=\int d\theta=\theta+C$

Since $x=\tan\theta, \theta=\tan^{-1}x$