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  4. If y= tan -1( (4x/1+5x2) )+ tan -1( (2+3x/3-2x) ), then (dy/dx) is equal to

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

KEAMKEAM 2008Continuity and Differentiability

Solution:

Given, $ y={{\tan }^{-1}}\left( \frac{4x}{1+5{{x}^{2}}} \right)+{{\tan }^{-1}}\left( \frac{2+3x}{3-2x} \right) $
Or $ y={{\tan }^{-1}}\left( \frac{5x-x}{1+5{{x}^{2}}} \right)+{{\tan }^{-1}}\left( \frac{\frac{2}{3}+x}{1-\frac{2}{3}x} \right) $
$ \Rightarrow $ $ y={{\tan }^{-1}}(5x)+{{\tan }^{-1}}(x) $ $ +{{\tan }^{-1}}\left( \frac{2}{3} \right)+{{\tan }^{-1}}(x) $
$ \Rightarrow $ $ y={{\tan }^{-1}}(5x)+{{\tan }^{-1}}\left( \frac{2}{3} \right) $ On differentiating w.r.t. $ x, $ we get $ \frac{dy}{dx}=\frac{5}{1+{{(5x)}^{2}}}=\frac{5}{1+25{{x}^{2}}} $