Hello,

in a physics exercise I need the derivative of $\displaystyle \mathrm{arctan}(x)$. 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 $\displaystyle - \frac{\pi}{2} < x < \frac{\pi}{2}$ strict monoton so that it is bijective and the inverse exists.

Now I calculate:

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

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

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

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

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

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

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

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

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

Could someone explain me the relation $\displaystyle \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