# successive differentiation

• Sep 15th 2012, 01:14 AM
prasum
successive differentiation
find the nth derivative of y=tan(inverse) ((x*sin a)/(1-x*cos a))
• Sep 15th 2012, 07:51 AM
Soroban
Re: successive differentiation
Hello, prasum!

The first derivative was hard enough.

Quote:

$\displaystyle \text{Find the }n^{th}\text{ derivative of: }\:y\;=\;\tan^{-1}\left(\frac{x\sin a}{1-x\cos a}\right)$

$\displaystyle \text{Let }\,\begin{Bmatrix}A = \sin a \\ B = \cos a \end{Bmatrix}\quad\text{ Note that: }\,A^2+B^2 \,=\,1$

We have: .$\displaystyle y \;=\;\tan^{-1}\left(\frac{Ax}{1-Bx}\right)$

$\displaystyle \frac{dy}{dx} \;=\;\frac{1}{1 + \left(\frac{Ax}{1-Bx}\right)^2} \cdot \frac{(1-Bx)A - Ax(-B)}{(1-Bx)^2}$

. . .$\displaystyle =\;\frac{(1-Bx)^2}{(1-Bx)^2 + (Ax)^2}\cdot \frac{A - ABx + ABx}{(1-Bx)^2} \;=\;\frac{A}{(1-Bx)^2 + (Ax)^2}$

. . .$\displaystyle =\;\frac{A}{1 - 2Bx + B^2x^2 + A^2x^2} \;=\;\frac{A}{1-2Bx + \underbrace{(A^2+B^2)}_{\text{This is 1}}x^2}$

$\displaystyle \frac{dy}{dx} \;=\;\frac{A}{1-2Bx + x^2} \;=\;A\big(1-2Bx+x^2\big)^{-1}$

Good luck!
• Sep 15th 2012, 07:56 AM
MarkFL
Re: successive differentiation
I did essentially the same thing as soroban, and even computed up the the 4th derivative, and only found a partial pattern...so I second the "Good luck!" :)