Prove $(\tan^{-1}{x})' = \frac{1}{1+x^2}$
How should I proceed? I tried:
Then another which I think is wrong:
What should I be doing?
UPDATE
In my lecture notes ... maybe a typo:
Prove $(\tan^{-1}{x})' = \frac{1}{1+x^2}$
How should I proceed? I tried:
Then another which I think is wrong:
What should I be doing?
UPDATE
In my lecture notes ... maybe a typo:
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}
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}
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
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 $x=\tan\theta, \theta=\tan^{-1}x$