If y = tan^{-1} left( frac{5x - x}{1 + 5x^2} | vedclass

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(B) Given $y = 	an^{-1} \left( \frac{5x - x}{1 + 5x^2} ight) + 	an^{-1} \left( \frac{2/3 + x}{1 - (2/3)x} ight)$.Using the formula $	an^{-1} a

Vedclass · Accepted Answer

$\frac{5}{1 + 25x^2}$

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(B) Given $y = 	an^{-1} \left( \frac{5x - x}{1 + 5x^2} ight) + 	an^{-1} \left( \frac{2/3 + x}{1 - (2/3)x} ight)$.Using the formula $	an^{-1} a - 	an^{-1} b = 	an^{-1} \left( \frac{a - b}{1 + ab} ight)$,the first term is $	an^{-1}(5x) - 	an^{-1}(x)$.Using the formula $	an^{-1} a + 	an^{-1} b = 	an^{-1} \left( \frac{a + b}{1 - ab} ight)$,the second term is $	an^{-1}(2/3) + 	an^{-1}(x)$.Thus,$y = 	an^{-1}(5x) - 	an^{-1}(x) + 	an^{-1}(2/3) + 	an^{-1}(x) = 	an^{-1}(5x) + 	an^{-1}(2/3)$.Differentiating with respect to $x$:$\frac{dy}{dx} = \frac{d}{dx} (	an^{-1}(5x)) + \frac{d}{dx} (	an^{-1}(2/3))$.Since $	an^{-1}(2/3)$ is a constant,its derivative is $0$.$\frac{dy}{dx} = \frac{1}{1 + (5x)^2} 	imes \frac{d}{dx}(5x) + 0 = \frac{5}{1 + 25x^2}$.

If $y = \tan^{-1} \left( \frac{5x - x}{1 + 5x^2} \right) + \tan^{-1} \left( \frac{2/3 + x}{1 - (2/3)x} \right)$,then $\frac{dy}{dx} =$

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