If y = tan^{-1}left(frac{4x}{1+5x^2}right) + | vedclass

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(B) Given $y = 	an^{-1}\left(\frac{4x}{1+5x^2}ight) + 	an^{-1}\left(\frac{3+8x}{8-3x}ight)$.We can rewrite the terms inside the inverse tangent

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

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(B) Given $y = 	an^{-1}\left(\frac{4x}{1+5x^2}ight) + 	an^{-1}\left(\frac{3+8x}{8-3x}ight)$.We can rewrite the terms inside the inverse tangent functions:$y = 	an^{-1}\left(\frac{5x-x}{1+5x \cdot x}ight) + 	an^{-1}\left(\frac{\frac{3}{8}+x}{1-\frac{3}{8} \cdot x}ight)$.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}(5x) - 	an^{-1}(x)) + (	an^{-1}(\frac{3}{8}) + 	an^{-1}(x))$.Simplifying the expression,we get:$y = 	an^{-1}(5x) + 	an^{-1}(\frac{3}{8})$.Now,differentiating with respect to $x$:$\frac{dy}{dx} = \frac{d}{dx}(	an^{-1}(5x)) + \frac{d}{dx}(	an^{-1}(\frac{3}{8}))$.Since $	an^{-1}(\frac{3}{8})$ is a constant,its derivative is $0$:$\frac{dy}{dx} = \frac{1}{1+(5x)^2} \cdot \frac{d}{dx}(5x) + 0$.$\frac{dy}{dx} = \frac{5}{1+25x^2}$.

If $y = \tan^{-1}\left(\frac{4x}{1+5x^2}\right) + \tan^{-1}\left(\frac{3+8x}{8-3x}\right)$,then $\frac{dy}{dx} = $

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