# Thread: first, second and third deriative of ln (tan x )

1. ## first, second and third deriative of ln (tan x )

y=ln(tan x), dy/dx=2/(sin2x) and d2y/dx2= -4(cos 2x)/(sin 2x)^2 , show that d3y/dx3 = ((4)(3+cos 4x))/(sin 2x)^3 ... i got the solution for dy/dx and d2y/dx2 but not d3y/dx3 , can anyone show me how to get d3y/dx3 please? Thanks in advance!

here's the working
https://www.flickr.com/photos/123101...3/13824984513/

2. ## Re: first, second and third deriative of ln (tan x )

Hello, delso!

$\displaystyle y\:=\:\ln(\tan x).\;\;\;\text{Find }\,\frac{dy}{dx},\;\frac{d^2y}{dx^2},\;\frac{d^3y} {dx^3}$

I would do it this way . . .

$\displaystyle y \;=\;\ln(\tan x)$

$\displaystyle \frac{dy}{dx} \;=\;\frac{1}{\tan x}\sec^2x \;=\;\frac{\cos x}{\sin x}\cdot\frac{1}{\cos^2x} \;=\;\frac{1}{\sin x\cos x}$

. . $\displaystyle =\;\frac{2}{2\sin x\cos x} \;=\;\frac{2}{\sin2x} \;=\;2\csc2x$

$\displaystyle \frac{d^2y}{dx^2} \;=\;2\cdot(-\csc2x\cot2x)\cdot2 \;=\;-4\csc2x\cot2x$

$\displaystyle \frac{d^3y}{dx^3} \;=\;-4\bigg[\csc2x(-\csc^22x\cdot 2) + \cot2x(-\csc2x\cot2x\cdot 2)\bigg]$

. . .$\displaystyle =\;8\:\!\csc2x(\csc^22x + \cot^22x)$

3. ## Re: first, second and third deriative of ln (tan x )

can you show that d3y/dx3 = ((4)(3+cos 4x))/(sin 2x)^3 ? i'm quite weak when it comes to trigonometry....thanks in advance!

4. ## Re: first, second and third deriative of ln (tan x )

Start by replacing $\displaystyle \cos{4x}$ with $\displaystyle \cos^2{2x}-\sin^2{2x}$ and 3 with $\displaystyle 3\cos^2{2x}+3\sin^2{2x}$.

- Hollywood