Question

proof of the nth derivative of arctan (x/a)

Answer 1

In the proof by induction , shown below, replace z = x/a.   and  follow this procedure..   dz/dx = 1/a.  substitute this and you will be able to prove that.  

$let\ sin\phi=\frac{a}{\sqrt{a^2+x^2}},\ \ cos\phi=\frac{x}{\sqrt{a^2+x^2}}\\\\ \frac{d\phi}{dx}= -\frac{1}{a}sin^2\phi\\\\<span>\frac{d^n}{dx^n}tan^{-1}\frac{x}{a}=\frac{(-1)^{n-1}(n-1)!}{\ (a^2+x^2)^{\frac{n}{2}}}Sin[n\ sin^{-1}(\frac{a}{\sqrt{a^2+x^2}})]$
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The formula for the n-th derivative of y = arctan (x), is proved by induction.

$\frac{d^n}{dx^n}tan^{-1}x=\frac{(-1)^{n-1}(n-1)!}{(1+x^2)^{\frac{n}{2}}}Sin[n\ sin^{-1}(\frac{1}{\sqrt{1+x^2}})],\\\\ for\ n=1,2,3,4,5.....\ eq\ (1)\\\\let\ sin\phi=\frac{1}{\sqrt{1+x^2}},\ \ so\ cos\phi=\frac{x}{\sqrt{1+x^2}},\\\\ cos\phi\ d\phi=-\frac{2x\ dx}{2(1+x^2)^\frac{3}{2}} \ \ \ \frac{d\phi}{dx}=-sin^2\phi \\\\ So\ (1) : y_x^n = (-1)^{n-1}(n-1)! \ Sin^n \phi\ Sin (n\phi)\ ....\ eq\ (2)$

Let (2) be true for n = k.

$Differentiate\ y^k_x\ wrt\ x.\\\\ y^{k+1}_x=(-1)^{k-1}(k-1)!\ [k\ sin^{k-1}\phi\ sin\ k\phi\ cos\phi + sin^k \phi\ k\ cos k\phi ] \frac{d\phi}{dx}\\\\=(-1)^{k-1} (k-1)! \ k\ sin^{k-1}\phi\ [sin\ k\phi\ cos\phi+sin\phi\ cos\ k\phi] [-sin^2\phi]\\\\=(-1)^k\ k!\ sin^{k+1}\phi\ sin[(k+1)\phi]$\

So the formula in (2) is true for n = k+1 also.    Hence, (1) is proved by mathematical Induction.
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$let\ z=\frac{x}{a},\ \ dx=a*dz\\\\y=tan^{-1}{z},\ \ y'_z=\frac{1}{1+z^2}\\\\y''_z=-\frac{2z}{(1+z^2)^2}\\\\y'''_z=-\frac{2(1+z^2)^2-2z*2(1+z^2)*2z}{(1+z^2)^4}=-\frac{2(1-3z^2)}{(1+z^2)^3}\\\\y^4_z=-2\frac{-(1+z^2)^3 *6z - (1-3z^2) 3(1+z^2)^2*2 z}{(1+z^2)^6}=\frac{2*6*z*2(1-z^2)}{(1+z^2)^4}$

replace z by x/a and you will get answer... and   use   dz = dx / a ..