successive differentiation

**prasum** · September 15th 2012, 01:14 AM

find the nth derivative of y=tan(inverse) ((x*sin a)/(1-x*cos a))

**Soroban** · September 15th 2012, 07:51 AM

Hello, prasum!

The first derivative was hard enough.

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

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

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

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

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

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

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

Good luck!

**MarkFL** · September 15th 2012, 07:56 AM

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!"

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