If y=tan ^{-1}left(frac{3+2 x}{2-3 x}right)+t | vedclass

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(B) Given $y = 	an ^{-1}\left(\frac{3+2 x}{2-3 x}ight)+	an ^{-1}\left(\frac{3 x}{1+4 x^2}ight)$.We can rewrite the first term by dividing the nu

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$\frac{4}{1+16 x^2}$

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(B) Given $y = 	an ^{-1}\left(\frac{3+2 x}{2-3 x}ight)+	an ^{-1}\left(\frac{3 x}{1+4 x^2}ight)$.We can rewrite the first term by dividing the numerator and denominator by $2$:$y = 	an ^{-1}\left(\frac{\frac{3}{2}+x}{1-\frac{3}{2} x}ight)+	an ^{-1}\left(\frac{4 x-x}{1+(4 x)(x)}ight)$Using the formula $	an ^{-1} A + 	an ^{-1} B = 	an ^{-1}\left(\frac{A+B}{1-AB}ight)$ and $	an ^{-1} A - 	an ^{-1} B = 	an ^{-1}\left(\frac{A-B}{1+AB}ight)$:$y = 	an ^{-1}\left(\frac{3}{2}ight) + 	an ^{-1} x + 	an ^{-1} (4 x) - 	an ^{-1} x$$y = 	an ^{-1}\left(\frac{3}{2}ight) + 	an ^{-1} (4 x)$Now,differentiating with respect to $x$:$\frac{d y}{d x} = 0 + \frac{1}{1+(4 x)^2} \cdot \frac{d}{d x}(4 x)$$\frac{d y}{d x} = \frac{4}{1+16 x^2}$

If $y=\tan ^{-1}\left(\frac{3+2 x}{2-3 x}\right)+\tan ^{-1}\left(\frac{3 x}{1+4 x^2}\right)$,then $\frac{d y}{d x}$ is equal to

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