ddx1btan-1xb-1btan-1xa=

y=1b tan-1 xb-1a tan-1 xa dydx=1b·11+xa2·1b-1a 11+xa2×1a =1x2+b2-1x2+a2 =a2-b2(x2+b2) (x2+a2)

ddx1btan-1xb-1btan-1xa=

y=1b tan-1 xb-1a tan-1 xa dydx=1b·11+xa2·1b-1a 11+xa2×1a =1x2+b2-1x2+a2 =a2-b2(x2+b2) (x2+a2)

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$\frac{d}{dx} [\frac{1}{b} \tan^{- 1} \frac{x}{b} - \frac{1}{b} \tan^{- 1} \frac{x}{a}] =$

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$\frac{1}{(x^{2} + a^{2}) (x^{2} + b^{2})}$

$\frac{x^{2} + a^{2}}{x^{2} + b^{2}}$

$\frac{a^{2} - b^{2}}{(x^{2} + a^{2}) (x^{2} + b^{2})}$

$\frac{b^{2} - a^{2}}{(x^{2} + a^{2}) (x^{2} + b^{2})}$

answer is C.

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Detailed Solution

$y = [\frac{1}{b} \tan^{- 1} \frac{x}{b} - \frac{1}{a} \tan^{- 1} \frac{x}{a}] \frac{dy}{dx} = \frac{1}{b} \cdot \frac{1}{1 + {(\frac{x}{a})}^{2}} \cdot \frac{1}{b} - \frac{1}{a} \frac{1}{1 + {(\frac{x}{a})}^{2}} \times \frac{1}{a} = \frac{1}{x^{2} + b^{2}} - \frac{1}{x^{2} + a^{2}} = \frac{a^{2} - b^{2}}{(x^{2} + b^{2}) (x^{2} + a^{2})}$