If y=tan-13x-x31-3x2+tan-14x-4x31-6x2+x then dydx=

y=tan-13x-x31-3x2+tan-14x-4x31-6x2+x4 =3tan-1x+4tan-1x =7tan-1x dydx=71+x2

If y=tan-13x-x31-3x2+tan-14x-4x31-6x2+x then dydx=

y=tan-13x-x31-3x2+tan-14x-4x31-6x2+x4 =3tan-1x+4tan-1x =7tan-1x dydx=71+x2

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If $y = \tan^{- 1} (\frac{3 x - x^{3}}{1 - 3 x^{2}}) + \tan^{- 1} (\frac{4 x - 4 x^{3}}{1 - 6 x^{2} + x})$ then $\frac{dy}{dx} =$

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

$\frac{6}{1 + x^{2}}$

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

$\frac{7}{1 + x^{2}}$

answer is D.

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Detailed Solution

$y = \tan^{- 1} (\frac{3 x - x^{3}}{1 - 3 x^{2}}) + \tan^{- 1} (\frac{4 x - 4 x^{3}}{1 - 6 x^{2} + x^{4}}) = 3 \tan^{- 1} x + 4 \tan^{- 1} x = 7 \tan^{- 1} x \frac{dy}{dx} = \frac{7}{1 + x^{2}}$

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If y=tan-13x-x31-3x2+tan-14x-4x31-6x2+x then dydx=

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If $y = \tan^{- 1} (\frac{3 x - x^{3}}{1 - 3 x^{2}}) + \tan^{- 1} (\frac{4 x - 4 x^{3}}{1 - 6 x^{2} + x})$ then $\frac{dy}{dx} =$