frac{d}{dx} left( tan^{-1} left( frac{x}{1+6x | vedclass

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(C) Let $y = 	an^{-1} \left( \frac{x}{1+6x^2} ight)$.We can rewrite the argument of $	an^{-1}$ as $\frac{3x - 2x}{1 + (3x)(2x)}$.Using the formula

Vedclass · Accepted Answer

$\frac{3}{1+9x^2} - \frac{2}{1+4x^2}$

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(C) Let $y = 	an^{-1} \left( \frac{x}{1+6x^2} ight)$.We can rewrite the argument of $	an^{-1}$ as $\frac{3x - 2x}{1 + (3x)(2x)}$.Using the formula $	an^{-1}(A) - 	an^{-1}(B) = 	an^{-1} \left( \frac{A-B}{1+AB} ight)$,we get:$y = 	an^{-1}(3x) - 	an^{-1}(2x)$.Now,differentiate with respect to $x$:$\frac{dy}{dx} = \frac{d}{dx} (	an^{-1}(3x)) - \frac{d}{dx} (	an^{-1}(2x))$.Using the chain rule $\frac{d}{dx} (	an^{-1}(u)) = \frac{1}{1+u^2} \cdot \frac{du}{dx}$:$\frac{dy}{dx} = \frac{1}{1+(3x)^2} \cdot 3 - \frac{1}{1+(2x)^2} \cdot 2$.$\frac{dy}{dx} = \frac{3}{1+9x^2} - \frac{2}{1+4x^2}$.Thus,the correct option is $C$.

$\frac{d}{dx} \left( \tan^{-1} \left( \frac{x}{1+6x^2} \right) \right) = $ . . . . . .

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