Find the derivative: frac{d}{dx} tan^{-1} lef | vedclass

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(D) Let $y = 	an^{-1} \left( \frac{ax - b}{bx + a} ight)$.We can rewrite the argument of the inverse tangent function by dividing the numerator and

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(D) Let $y = 	an^{-1} \left( \frac{ax - b}{bx + a} ight)$.We can rewrite the argument of 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}(A) - 	an^{-1}(B) = 	an^{-1} \left( \frac{A - B}{1 + AB} ight)$,we get:$y = 	an^{-1}(x) - 	an^{-1} \left( \frac{b}{a} ight)$.Now,differentiate with respect to $x$:$\frac{dy}{dx} = \frac{d}{dx} \left( 	an^{-1}(x) ight) - \frac{d}{dx} \left( 	an^{-1} \left( \frac{b}{a} ight) ight)$.Since $	an^{-1} \left( \frac{b}{a} ight)$ is a constant,its derivative is $0$.Therefore,$\frac{dy}{dx} = \frac{1}{1 + x^2} - 0 = \frac{1}{1 + x^2}$.Comparing this with the given options,the correct answer is not listed.

Find the derivative: $\frac{d}{dx} \tan^{-1} \left( \frac{ax - b}{bx + a} \right) = $

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