# Thread: Derivative of Arctan

1. ## Derivative of Arctan

Hello,

in a physics exercise I need the derivative of $\mathrm{arctan}(x)$. I don't want not only look in my Bronstein for the derivative, I want to calculate it on my own by using the theorem of the inverse function. The tangent is in the intervall $- \frac{\pi}{2} < x < \frac{\pi}{2}$ strict monoton so that it is bijective and the inverse exists.

Now I calculate:

$y = f(x) = \mathrm{arctan}(x)$

$x = g(y) = \mathrm{tan} (y)$

$f'(x) \cdot g'(y) = 1$

$\Rightarrow f'(x) = \frac{1}{g'(y)}$

$g'(y) = \frac{\mathrm{d}}{\mathrm{d}y} ~ \frac{\mathrm{sin}(y)}{\mathrm{cos}(y)} \stackrel{\text{quotient rule}}= \frac{1}{\mathrm{cos}^{2}(y)}$

Now I have the problem, that I don't know how to express this in $x$ for a clear correspondence. I looked in my mathbook and found the relation:

$\frac{1}{\mathrm{cos}^{2}(y)} = 1 + \mathrm{tan}^{2}(y)$

With this it is clear, because $\mathrm{tan}(y) = x$ according to assumption and I got:

$f'(x) = \frac{1}{1+ \mathrm{tan}^{2}(y)} = \frac{1}{1+ x^{2}}$.

Could someone explain me the relation $\frac{1}{\mathrm{cos}^{2}(y)} = 1 + \mathrm{tan}^{2}(y)$ please? I thought about trigonometric relations like pythagoras but I did not see it.

Thanks for help

2. Originally Posted by Besserwisser
Hello,

in a physics exercise I need the derivative of $\mathrm{arctan}(x)$. I don't want not only look in my Bronstein for the derivative, I want to calculate it on my own by using the theorem of the inverse function. The tangent is in the intervall $- \frac{\pi}{2} < x < \frac{\pi}{2}$ strict monoton so that it is bijective and the inverse exists.

Now I calculate:

$y = f(x) = \mathrm{arctan}(x)$

$x = g(y) = \mathrm{tan} (y)$

$f'(x) \cdot g'(y) = 1$

$\Rightarrow f'(x) = \frac{1}{g'(y)}$

$g'(y) = \frac{\mathrm{d}}{\mathrm{d}y} ~ \frac{\mathrm{sin}(y)}{\mathrm{cos}(y)} \stackrel{\text{quotient rule}}= \frac{1}{\mathrm{cos}^{2}(y)}$

Now I have the problem, that I don't know how to express this in $x$ for a clear correspondence. I looked in my mathbook and found the relation:

$\frac{1}{\mathrm{cos}^{2}(y)} = 1 + \mathrm{tan}^{2}(y)$

With this it is clear, because $\mathrm{tan}(y) = x$ according to assumption and I got:

$f'(x) = \frac{1}{1+ \mathrm{tan}^{2}(y)} = \frac{1}{1+ x^{2}}$.

Could someone explain me the relation $\frac{1}{\mathrm{cos}^{2}(y)} = 1 + \mathrm{tan}^{2}(y)$ please? I thought about trigonometric relations like pythagoras but I did not see it.

Thanks for help

What you did is correct, and the explanation is simple trigonometry:

$1+\tan^2x=1+\frac{\sin^2x}{\cos^2x}=\frac{\cos^2x+ \sin^2x}{\cos^2x}=\frac{1}{\cos^2x}$ , using the trigonometric Pythagoras Theorem $\cos^2x+\sin^2x=1$

Tonio

3. Just in case a picture helps...

... where

... is the chain rule. Straight continuous lines differentiate downwards (integrate up) with respect to x, and the straight dashed line similarly but with respect to the dashed balloon expression (the inner function of the composite which is subject to the chain rule).

Follow clockwise from top left.

__________________________________________

Don't integrate - balloontegrate!

Balloon Calculus; standard integrals, derivatives and methods

Balloon Calculus Drawing with LaTeX and Asymptote!

4. ## thanks

Originally Posted by tonio

$1+\tan^2x=1+\frac{\sin^2x}{\cos^2x}=\frac{\cos^2x+ \sin^2x}{\cos^2x}=\frac{1}{\cos^2x}$
Oh, I did not see the forrest because of all the trees

Thank you very much. Now it is clear

Greetings