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

Question

Isaiah Allen · Accepted Answer

Remembering the derivative of #arctanx# 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.

#d/(dx)[arctanu] = 1/(1+u^2)((du)/(dx))#

Thus, we have:

#y = arctan(secx + tanx)#

#(dy)/(dx) = 1/(1+(secx + tanx)^2)(secxtanx + sec^2x)#

#= (secxtanx + sec^2x)/(1+(secx + tanx)^2)#

#= (secxtanx + sec^2x)/(1+sec^2x + 2secxtanx + tan^2x)#

#= (sec^2x + secxtanx)/(2sec^2x + 2secxtanx)#

#= color(blue)(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.

#tany = secx + tanx#

#sec^2y * (dy)/(dx) = secxtanx + sec^2x#

#(1 + tan^2color(green)(y)) * (dy)/(dx) = secxtanx + sec^2x#

Note that you have to use this identity, otherwise you'll get #sec^2(arctan(secx + tanx))#, which is not easy to work with.

#[1 + (tan[color(green)(arctan(secx + tanx))])^2] (dy)/(dx) = secxtanx + sec^2x#

#[1 + (secx + tanx)^2] (dy)/(dx) = secxtanx + sec^2x#

#(dy)/(dx) = (secxtanx + sec^2x)/(1 + (secx + tanx)^2)#

which, from above, is:

#color(blue)(= 1/2)#

Ezra Adams · Answer

undefined To find the derivative of $ y = \arctan(\sec x + \tan x) $, we'll first need to use the chain rule. The derivative of $ \arctan(u) $ with respect to $ u $ is $ \frac{1}{1 + u^2} $, and then we'll multiply by the derivative of the inside function.

So, let's differentiate $ \sec x + \tan x $ with respect to $ x $. The derivative of $ \sec x $ is $ \sec x \tan x $, and the derivative of $ \tan x $ is $ \sec^2 x $.

Putting it all together:

$$
\begin{align*}
\frac{d}{dx} \left( \arctan(\sec x + \tan x) \right) &= \frac{1}{1 + (\sec x + \tan x)^2} \cdot \left( \frac{d}{dx} (\sec x + \tan x) \right) \
&= \frac{1}{1 + (\sec x + \tan x)^2} \cdot (\sec x \tan x + \sec^2 x)
\end{align*}
$$