Question

D/dx tan⁻¹(x+a/1-xa)=.......... (x, a ∈ R⁺, xa > 1) ,​

Answer 1

$\displaystyle\huge\red{\underline{\underline{Solution}}}$

FORMULA TO BE IMPLEMENTED

1. $\displaystyle \: {tan}^{ - 1} ( \frac{x + a}{1 - xa} \: ) = {tan}^{ - 1}x + {tan}^{ - 1}a$

2.$\displaystyle \: \frac{d}{dx} ({tan}^{ - 1}x) = \frac{1}{1 + {x}^{2} }$

3. $\displaystyle \: \frac{d}{dx} ( c) = 0$

TO DETERMINE

$\displaystyle \: \frac{d}{dx} \{ {tan}^{ - 1} ( \frac{x + a}{1 - xa} \: ) \}$

CALCULATION

$\displaystyle \: \frac{d}{dx} \{ {tan}^{ - 1} ( \frac{x + a}{1 - xa} \: ) \}$

$= \displaystyle \: \frac{d}{dx} ( {tan}^{ - 1}x + {tan}^{ - 1}a) \:$ (BY FORMULA 1)

$= \displaystyle \: \frac{d}{dx} ( {tan}^{ - 1}x )+ \frac{d}{dx} ({tan}^{ - 1}a) \:$

$= \displaystyle \: \frac{1}{1 + {x}^{2} } + 0$ ( BY FORMULA 2 & 3)

$= \displaystyle \: \frac{1}{1 + {x}^{2} }$