# What is the derivative of y=arctan(secx + tanx)?

Aug 16, 2015

Remembering the derivative of $\arctan x$ here is very helpful. If you don't, it could get a bit frustrating with implicit differentiation. I'll do it both ways to show you what I mean.

$\frac{d}{\mathrm{dx}} \left[\arctan u\right] = \frac{1}{1 + {u}^{2}} \left(\frac{\mathrm{du}}{\mathrm{dx}}\right)$

Thus, we have:

$y = \arctan \left(\sec x + \tan x\right)$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{1 + {\left(\sec x + \tan x\right)}^{2}} \left(\sec x \tan x + {\sec}^{2} x\right)$

$= \frac{\sec x \tan x + {\sec}^{2} x}{1 + {\left(\sec x + \tan x\right)}^{2}}$

$= \frac{\sec x \tan x + {\sec}^{2} x}{1 + {\sec}^{2} x + 2 \sec x \tan x + {\tan}^{2} x}$

$= \frac{{\sec}^{2} x + \sec x \tan x}{2 {\sec}^{2} x + 2 \sec x \tan x}$

$= \textcolor{b l u e}{\frac{1}{2}}$

(Haha, nice. A derivative that doesn't even integrate back into the original function without some special manipulation.)

And now the other way.

$\tan y = \sec x + \tan x$

${\sec}^{2} y \cdot \frac{\mathrm{dy}}{\mathrm{dx}} = \sec x \tan x + {\sec}^{2} x$

$\left(1 + {\tan}^{2} \textcolor{g r e e n}{y}\right) \cdot \frac{\mathrm{dy}}{\mathrm{dx}} = \sec x \tan x + {\sec}^{2} x$

Note that you have to use this identity, otherwise you'll get ${\sec}^{2} \left(\arctan \left(\sec x + \tan x\right)\right)$, which is not easy to work with.

$\left[1 + {\left(\tan \left[\textcolor{g r e e n}{\arctan \left(\sec x + \tan x\right)}\right]\right)}^{2}\right] \frac{\mathrm{dy}}{\mathrm{dx}} = \sec x \tan x + {\sec}^{2} x$

$\left[1 + {\left(\sec x + \tan x\right)}^{2}\right] \frac{\mathrm{dy}}{\mathrm{dx}} = \sec x \tan x + {\sec}^{2} x$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\sec x \tan x + {\sec}^{2} x}{1 + {\left(\sec x + \tan x\right)}^{2}}$

which, from above, is:

$\textcolor{b l u e}{= \frac{1}{2}}$