If y = tan^{-1} left{ frac{ax - b}{bx + a} ri | vedclass

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(B) Given $y = 	an^{-1} \left\{ \frac{ax - b}{bx + a} ight\}$.We can rewrite the expression inside the inverse tangent function by dividing the num

Vedclass · Accepted Answer

$\frac{1}{1 + x^2}$

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(B) Given $y = 	an^{-1} \left\{ \frac{ax - b}{bx + a} ight\}$.We can rewrite the expression inside the inverse tangent function by dividing the numerator and denominator by $a$:$y = 	an^{-1} \left\{ \frac{x - \frac{b}{a}}{1 + \frac{b}{a}x} ight\}$.Using the identity $	an^{-1} \left( \frac{A - B}{1 + AB} ight) = 	an^{-1} A - 	an^{-1} B$,we get:$y = 	an^{-1} x - 	an^{-1} \left( \frac{b}{a} ight)$.Now,differentiating with respect to $x$:$y' = \frac{d}{dx} (	an^{-1} x) - \frac{d}{dx} (	an^{-1} \frac{b}{a})$.Since $	an^{-1} \left( \frac{b}{a} ight)$ is a constant,its derivative is $0$.Therefore,$y' = \frac{1}{1 + x^2} - 0 = \frac{1}{1 + x^2}$.

If $y = \tan^{-1} \left\{ \frac{ax - b}{bx + a} \right\}$,then $y' = $

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