# Question 84eb0

##### 1 Answer
Jan 24, 2018

$f ' \left(x\right) = \frac{\sqrt{1 - \tan x}}{\left(1 - \sin \left(2 x\right)\right) \sqrt{1 + \tan x}}$

#### Explanation:

Given: $f \left(x\right) = \sqrt{\frac{1 + \tan x}{1 - \tan x}}$

To make this easier visually, we'll let $\frac{1 + \tan x}{1 - \tan x} = u$

So we have $f \left(x\right) = \sqrt{u}$

By the chain rule we're solving this:

color(red)(d/(du)[sqrtu])*color(blue)(d/dx[(1+tanx)/(1-tanx)]

Each derivative is a different technique so that's why a color-coded it.

color(red)(d/dx[sqrtu]=>d/dx[u^(1/2)]

$\textcolor{red}{\text{By the Power Rule}}$

color(red)(=1/2u^(1/2-1)=1/2u^(-1/2)=1/(2u^(1/2))=1/(2sqrtu)

color(blue)(d/dx[(1+tanx)/(1-tanx)]larr "This requires the quotient rule"

color(blue)(=(d/dx[1+tanx]*(1-tanx)-d/dx[1-tanx]*(1+tanx))/(1-tanx)^2

color(blue)(=(sec^2x*(1-tanx)--sec^2x*(1+tanx))/(1-tanx)^2

color(blue)("We can simplify this if we rewrite everything in terms of sines and cosines"

color(blue)(=(1/cos^2x*(1-sinx/cosx)+1/cos^2x*(1+sinx/cosx))/(1-sinx/cosx)^2

color(blue)("Combine " (1-sinx/cosx) & (1+sinx/cosx)" And Simplify"

color(blue)(=(1/cos^2x*((cosx-sinx)/cosx)+1/cos^2x*((cosx+sinx)/cosx))/((cosx-sinx)/cosx)^2

=color(blue)(((cosx-sinx)/cos^3x)+((cosx+sinx)/cos^3x))/(color(blue)((cosx-sinx)^2/cos^2x)

color(blue)("Combine the numerator and simplify"

color(blue)(=((cosx-sinx+cosx+sinx)/cos^3x)/((cosx-sinx)^2/cos^2x)

color(blue)(=((2cosx)/cos^3x)/((cosx-sinx)^2/cos^2x)

color(blue)(=(2/cos^2x)/((cosx-sinx)^2/cos^2x)

color(blue)("Divide Fractions " (a/b)/(c/d)=(a*d)/(b*c) 

color(blue)(=2/cos^2x*cos^2x/(cosx-sinx)^2=(2cos^2x)/(cos^2x(cosx-sinx)^2)

color(blue)(=(2)/((cosx-sinx)^2)

color(blue)("Expand " (cosx-sinx)^2 " and rewrite"

color(blue)(=2/(cos^2x-2cosxsinx+sin^2x)

color(blue)(=2/(cos^2x+sin^2x-2cosxsinx)

color(blue)("Use the identities and rewrite":

color(blue)(cos^2x+sin^2x=1

color(blue)(2cosxsinx=sin(2x)

color(blue)("So we get.."

color(blue)(=2/(1-sin(2x))

So now that we have our two derivatives we can put it all together to get:

f'(x)=color(red)(1/(2sqrtu))*color(blue)(2/(1-sin(2x))#

Recall that we have to reverse the substitution and substitute back $\frac{1 + \tan x}{1 - \tan x}$ for $u$

$f ' \left(x\right) = \frac{1}{2 \sqrt{\frac{1 + \tan x}{1 - \tan x}}} \cdot \frac{2}{1 - \sin \left(2 x\right)}$

We can simplify this if need be but I'm pretty tired from all the work done previously but you will get:

$f ' \left(x\right) = \frac{\sqrt{1 - \tan x}}{\left(1 - \sin \left(2 x\right)\right) \sqrt{1 + \tan x}}$

I think a very important lesson here (seeing how this is a lot of work) is that don't be discouraged by the size of the problem. If you think about it, this was more algebra/trigonometry than calculus. In addition, you must be on top of trigonometry skills because they will show up everywhere in calculus as seen here.
So work hard and stay sharp! :)