Question

proof of nth derivative of tan inverse x.

Answer 1

y = Tan⁻¹ (x/a)

The formula for nth derivative is proved by Mathematical induction process.
Formula:

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.  

 

The formula for the n-th derivative of y = arctan (x), is proved by induction.



Let (2) be true for n = k.



So the formula in (2) is true for n = k+1 also.    Hence, (1) is proved by mathematical Induction.

Answer 2

$n^{th}$ derivative of $tan^{-1} x$ is $\frac{1}{2i}(-1)^{n-1} (n-1)![(x-i)^{-n} -(x+i)^{-n}]$

Step-by-step explanation:

Given,

$n^{th}$ derivative of $tan^{-1} x$

∴ $y=tan^{-1} x$

Differentiate above equation with respect to $x$,

⇒$y_{1} =\frac{d}{dx}( tan^{-1} x)$  where $y_{1}=\frac{dy}{dx}$

⇒$y_{1} =\frac{1}{1+x^{2} }$ (∵ $\frac{d}{dx}( tan^{-1} x=\frac{1}{1+x^{2} }$

⇒$y_{1}=\frac{1}{x^{2}-i^{2}  }$ (∵ $i^{2} =-1$)

⇒$y_{1}=\frac{1}{(x+i)(x-i)}$

⇒$y_{1}=\frac{1}{2i} [\frac{(x+i)-(x-i)}{(x+i)(x-i)}]$

⇒$y_{1}=\frac{1}{2i}[\frac{1}{x-i} -\frac{1}{x+i} ]$

⇒$y_{1}=\frac{1}{2i}[(x-i)^{-1} -(x+i)^{-1}]$

Then, $2^{nd}$ derivative,

$y_{2}=\frac{1}{2i}[(-1)(x-i)^{-2} -(-1)(x+i)^{-2}]$

Similarly, $3^{rd}$ derivative,

$y_{3}=\frac{1}{2i}[(-1)(-2)(x-i)^{-3} -(-1)(-2)(x+i)^{-3}]$

⇒$y_{3}=\frac{1}{2i}(-1)^{2} (2!)[(x-i)^{-3} -(x+i)^{-3}]$

Similarly, $n^{th}$ derivative,

∴ $y_{n}=\frac{1}{2i}(-1)^{n-1} (n-1)![(x-i)^{-n} -(x+i)^{-n}]$ where $y_{n}=\frac{d^{n}y }{dx^{n} }$

So, $n^{th}$ derivative of $tan^{-1} x$ is $\frac{1}{2i}(-1)^{n-1} (n-1)![(x-i)^{-n} -(x+i)^{-n}]$